Complex Numbers - 2

N. B. Vyas

Department of Mathematics,Atmiya Institute of Tech. and Science,

Rajkot (Guj.)

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ

inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger

(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ

+ nC1 cosn−1θ (isinθ) +

nC2 cosn−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ)

+nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 +

. . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)

= ( nC0 cosnθ − nC2 cos

n−2θ sin2θ + . . .)+ i ( nC1 cos

n−1θ sinθ − nC3 cosn−3θ sin3θ + . . .)

By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ

− nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ

− nC3 cosn−3θ sin3θ + . . .)

By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)

By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we get

cos nθ = nC0 cosnθ − nC2 cos

n−2θ sin2θ + . . .sin nθ = nC1 cos

n−1θ sinθ− nC3 cosn−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sin nθ and Cos nθ inpowers of sinθ cosθ, where n is a positiveinteger(cosnθ + isinnθ) = (cosθ + isinθ)n

= nC0 cosnθ + nC1 cos

n−1θ (isinθ) +nC2 cos

n−2θ (isinθ)2 + . . .+ nCn (isinθ)n

(Using Binomial theorem)= ( nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .)

+ i ( nC1 cosn−1θ sinθ − nC3 cos

n−3θ sin3θ + . . .)By comparing real and imaginary parts on bothsides, we getcos nθ = nC0 cos

nθ − nC2 cosn−2θ sin2θ + . . .

sin nθ = nC1 cosn−1θ sinθ− nC3 cos

n−3θ sin3θ+ . . .N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then

zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ

∴1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn=

cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z=

2cosθ ⇒ cosθ =1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)

and zn +1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn=

2cos nθ ⇒ cos nθ =1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)

cosnθ =1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)n

similarly sinθ =1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)

sinn θ =1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)n

N. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)nN. B. Vyas Complex Numbers - 2

Expansion Using De Moivre’s Theorem

Method of Expansion of Sinnθ and Cosnθ

Let z = cosθ + isinθ then zn = cos nθ + i sin nθ

Also1

z= cosθ − isinθ ∴

1

zn= cos nθ − i sin nθ

∴ z +1

z= 2cosθ ⇒ cosθ =

1

2

(z +

1

z

)and zn +

1

zn= 2cos nθ ⇒ cos nθ =

1

2

(zn +

1

zn

)cosnθ =

1

2n

(z +

1

z

)nsimilarly sinθ =

1

2i

(z − 1

z

)and sin nθ =

1

2i

(zn − 1

zn

)sinn θ =

1

(2i)n

(z − 1

z

)nN. B. Vyas Complex Numbers - 2

Examples

Ex. Prove that cos4θ = cos4θ − 6cos2θsin2θ + sin4θ

N. B. Vyas Complex Numbers - 2

Examples

Ex. Using De Moivre’s theorem prove the following:

(i) cos6θ = 32 cos6θ − 48 cos4θ + 18 cos2θ − 1

(ii) sin6θ = 3 sin2θ − 4 sin32θ

(iii) tan6θ =6 tanθ − 20 tan3θ + 6 tan5θ

1− 15 tan2θ + 15 tan4θ − tan6θ

N. B. Vyas Complex Numbers - 2

Examples

Ex. Using De Moivre’s theorem prove the following:

(i) cos6θ = 32 cos6θ − 48 cos4θ + 18 cos2θ − 1

(ii) sin6θ = 3 sin2θ − 4 sin32θ

(iii) tan6θ =6 tanθ − 20 tan3θ + 6 tan5θ

1− 15 tan2θ + 15 tan4θ − tan6θ

N. B. Vyas Complex Numbers - 2

Examples

Ex. Using De Moivre’s theorem prove the following:

(i) cos6θ = 32 cos6θ − 48 cos4θ + 18 cos2θ − 1

(ii) sin6θ = 3 sin2θ − 4 sin32θ

(iii) tan6θ =6 tanθ − 20 tan3θ + 6 tan5θ

1− 15 tan2θ + 15 tan4θ − tan6θ

N. B. Vyas Complex Numbers - 2

Examples

Ex. Using De Moivre’s theorem prove the following:

(i) cos6θ = 32 cos6θ − 48 cos4θ + 18 cos2θ − 1

(ii) sin6θ = 3 sin2θ − 4 sin32θ

(iii) tan6θ =6 tanθ − 20 tan3θ + 6 tan5θ

1− 15 tan2θ + 15 tan4θ − tan6θ

N. B. Vyas Complex Numbers - 2

Examples

Ex. Prove that

cos8θ =1

27(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 +

8C1 z7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+

8C2 z6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+

8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+

8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 +

8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 +

28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 +

56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 +

70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

Sol.: We know that,

(2cosθ)8 =

(z +

1

z

)8

= 8C0 z8 + 8C1 z

7

(1

z

)+ 8C2 z

6

(1

z

)2

+

8C3 z5

(1

z

)3

+ 8C4 z4

(1

z

)4

+ 8C5 z3

(1

z

)5

+

8C6 z2

(1

z

)6

+ 8C7 z

(1

z

)7

+ 8C8

(1

z

)8

= z8 + 8z6 + 28z4 + 56z2 + 70 +56

z2+

28

z4+

8

z6+

1

z8

N. B. Vyas Complex Numbers - 2

Examples

=

(z8 +

1

z8

)+

8

(z6 +

1

z6

)+ 28

(z4 +

1

z4

)+

56

(z2 +

1

z2

)+ 70

= 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + 70

∴ cos8θ =1

27(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Examples

=

(z8 +

1

z8

)+ 8

(z6 +

1

z6

)+

28

(z4 +

1

z4

)+

56

(z2 +

1

z2

)+ 70

= 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + 70

∴ cos8θ =1

27(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Examples

=

(z8 +

1

z8

)+ 8

(z6 +

1

z6

)+ 28

(z4 +

1

z4

)+

56

(z2 +

1

z2

)+ 70

= 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + 70

∴ cos8θ =1

27(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Examples

=

(z8 +

1

z8

)+ 8

(z6 +

1

z6

)+ 28

(z4 +

1

z4

)+

56

(z2 +

1

z2

)+ 70

= 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + 70

∴ cos8θ =1

27(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Examples

=

(z8 +

1

z8

)+ 8

(z6 +

1

z6

)+ 28

(z4 +

1

z4

)+

56

(z2 +

1

z2

)+ 70

= 2cos8θ +

16cos6θ + 56cos4θ + 112cos2θ + 70

∴ cos8θ =1

27(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Examples

=

(z8 +

1

z8

)+ 8

(z6 +

1

z6

)+ 28

(z4 +

1

z4

)+

56

(z2 +

1

z2

)+ 70

= 2cos8θ + 16cos6θ +

56cos4θ + 112cos2θ + 70

∴ cos8θ =1

27(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Examples

=

(z8 +

1

z8

)+ 8

(z6 +

1

z6

)+ 28

(z4 +

1

z4

)+

56

(z2 +

1

z2

)+ 70

= 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ +

70

∴ cos8θ =1

27(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Examples

=

(z8 +

1

z8

)+ 8

(z6 +

1

z6

)+ 28

(z4 +

1

z4

)+

56

(z2 +

1

z2

)+ 70

= 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + 70

∴ cos8θ =1

27(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Examples

=

(z8 +

1

z8

)+ 8

(z6 +

1

z6

)+ 28

(z4 +

1

z4

)+

56

(z2 +

1

z2

)+ 70

= 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + 70

∴ cos8θ =1

27

(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Examples

=

(z8 +

1

z8

)+ 8

(z6 +

1

z6

)+ 28

(z4 +

1

z4

)+

56

(z2 +

1

z2

)+ 70

= 2cos8θ + 16cos6θ + 56cos4θ + 112cos2θ + 70

∴ cos8θ =1

27(cos8θ+8cos6θ+28cos4θ+56cos2θ+35)

N. B. Vyas Complex Numbers - 2

Circular Functions of a Complex Numbers

eix = cosx+ isinx

and e−ix = cosx− isinxare known as Euler’s formula.

By adding, we get them, we get

cosx =eix + e−ix

2and sinx =

eix − e−ix

2iare known as circular functions and are true forall real values of x.

They are also known as Euler’s exponentialform of cosines and sines.

N. B. Vyas Complex Numbers - 2

Circular Functions of a Complex Numbers

eix = cosx+ isinx and e−ix = cosx− isinx

are known as Euler’s formula.

By adding, we get them, we get

cosx =eix + e−ix

2and sinx =

eix − e−ix

2iare known as circular functions and are true forall real values of x.

They are also known as Euler’s exponentialform of cosines and sines.

N. B. Vyas Complex Numbers - 2

Circular Functions of a Complex Numbers

eix = cosx+ isinx and e−ix = cosx− isinxare known as Euler’s formula.

By adding, we get them, we get

cosx =eix + e−ix

2and sinx =

eix − e−ix

2iare known as circular functions and are true forall real values of x.

They are also known as Euler’s exponentialform of cosines and sines.

N. B. Vyas Complex Numbers - 2

Circular Functions of a Complex Numbers

eix = cosx+ isinx and e−ix = cosx− isinxare known as Euler’s formula.

By adding, we get them, we get

cosx =eix + e−ix

2and sinx =

eix − e−ix

2iare known as circular functions and are true forall real values of x.

They are also known as Euler’s exponentialform of cosines and sines.

N. B. Vyas Complex Numbers - 2

Circular Functions of a Complex Numbers

eix = cosx+ isinx and e−ix = cosx− isinxare known as Euler’s formula.

By adding, we get them, we get

cosx =eix + e−ix

2

and sinx =eix − e−ix

2iare known as circular functions and are true forall real values of x.

They are also known as Euler’s exponentialform of cosines and sines.

N. B. Vyas Complex Numbers - 2

Circular Functions of a Complex Numbers

eix = cosx+ isinx and e−ix = cosx− isinxare known as Euler’s formula.

By adding, we get them, we get

cosx =eix + e−ix

2and sinx =

eix − e−ix

2i

are known as circular functions and are true forall real values of x.

They are also known as Euler’s exponentialform of cosines and sines.

N. B. Vyas Complex Numbers - 2

Circular Functions of a Complex Numbers

eix = cosx+ isinx and e−ix = cosx− isinxare known as Euler’s formula.

By adding, we get them, we get

cosx =eix + e−ix

2and sinx =

eix − e−ix

2iare known as circular functions and are true forall real values of x.

They are also known as Euler’s exponentialform of cosines and sines.

N. B. Vyas Complex Numbers - 2

Circular Functions of a Complex Numbers

eix = cosx+ isinx and e−ix = cosx− isinxare known as Euler’s formula.

By adding, we get them, we get

cosx =eix + e−ix

2and sinx =

eix − e−ix

2iare known as circular functions and are true forall real values of x.

They are also known as Euler’s exponentialform of cosines and sines.

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The hyperbolic sine of x is denoted by sinh(x),

the hyperbolic cosine of x is defined by cosh(x)and hyperbolic tangent of x is defined by tanh(x)and are defined respectively as

sinh(x) =ex − e−x

2

cosh(x) =ex + e−x

2

and tanh(x) =ex − e−x

ex + e−x;xεR

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The hyperbolic sine of x is denoted by sinh(x),the hyperbolic cosine of x is defined by cosh(x)

and hyperbolic tangent of x is defined by tanh(x)and are defined respectively as

sinh(x) =ex − e−x

2

cosh(x) =ex + e−x

2

and tanh(x) =ex − e−x

ex + e−x;xεR

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The hyperbolic sine of x is denoted by sinh(x),the hyperbolic cosine of x is defined by cosh(x)and hyperbolic tangent of x is defined by tanh(x)

and are defined respectively as

sinh(x) =ex − e−x

2

cosh(x) =ex + e−x

2

and tanh(x) =ex − e−x

ex + e−x;xεR

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The hyperbolic sine of x is denoted by sinh(x),the hyperbolic cosine of x is defined by cosh(x)and hyperbolic tangent of x is defined by tanh(x)and are defined respectively as

sinh(x) =ex − e−x

2

cosh(x) =ex + e−x

2

and tanh(x) =ex − e−x

ex + e−x;xεR

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The hyperbolic sine of x is denoted by sinh(x),the hyperbolic cosine of x is defined by cosh(x)and hyperbolic tangent of x is defined by tanh(x)and are defined respectively as

sinh(x) =ex − e−x

2

cosh(x) =ex + e−x

2

and tanh(x) =ex − e−x

ex + e−x;xεR

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The hyperbolic sine of x is denoted by sinh(x),the hyperbolic cosine of x is defined by cosh(x)and hyperbolic tangent of x is defined by tanh(x)and are defined respectively as

sinh(x) =ex − e−x

2

cosh(x) =ex + e−x

2

and tanh(x) =ex − e−x

ex + e−x;xεR

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The hyperbolic sine of x is denoted by sinh(x),the hyperbolic cosine of x is defined by cosh(x)and hyperbolic tangent of x is defined by tanh(x)and are defined respectively as

sinh(x) =ex − e−x

2

cosh(x) =ex + e−x

2

and tanh(x) =ex − e−x

ex + e−x;xεR

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The reciprocals of these functions are defined asbelow

cosech(x) =2

ex − e−x

sech(x) =2

ex − e−x

and coth(x) =ex + e−x

ex − e−x

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The reciprocals of these functions are defined asbelow

cosech(x) =2

ex − e−x

sech(x) =2

ex − e−x

and coth(x) =ex + e−x

ex − e−x

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The reciprocals of these functions are defined asbelow

cosech(x) =2

ex − e−x

sech(x) =2

ex − e−x

and coth(x) =ex + e−x

ex − e−x

N. B. Vyas Complex Numbers - 2

Hyperbolic functions

The reciprocals of these functions are defined asbelow

cosech(x) =2

ex − e−x

sech(x) =2

ex − e−x

and coth(x) =ex + e−x

ex − e−x

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

We know that sinx =eix − e−ix

2i

replacing x by ix

sin(ix) =ei(ix) − e−i(ix)

2i=e−x − ex

2i

= −ex − e−x

2i= i

(ex − e−x

2

)∴ sin(ix) = isinh(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

We know that sinx =eix − e−ix

2ireplacing x by ix

sin(ix) =ei(ix) − e−i(ix)

2i=e−x − ex

2i

= −ex − e−x

2i= i

(ex − e−x

2

)∴ sin(ix) = isinh(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

We know that sinx =eix − e−ix

2ireplacing x by ix

sin(ix) =ei(ix) − e−i(ix)

2i=e−x − ex

2i

= −ex − e−x

2i= i

(ex − e−x

2

)∴ sin(ix) = isinh(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

We know that sinx =eix − e−ix

2ireplacing x by ix

sin(ix) =ei(ix) − e−i(ix)

2i=e−x − ex

2i

= −ex − e−x

2i= i

(ex − e−x

2

)

∴ sin(ix) = isinh(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

We know that sinx =eix − e−ix

2ireplacing x by ix

sin(ix) =ei(ix) − e−i(ix)

2i=e−x − ex

2i

= −ex − e−x

2i= i

(ex − e−x

2

)∴ sin(ix) = isinh(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cos(ix) = cosh(x)

tan(ix) = i tanh(x)

cosec(ix) = −i cosech(x)

sec(ix) = sech(x)

cot(ix) = −i coth(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cos(ix) = cosh(x)

tan(ix) = i tanh(x)

cosec(ix) = −i cosech(x)

sec(ix) = sech(x)

cot(ix) = −i coth(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cos(ix) = cosh(x)

tan(ix) = i tanh(x)

cosec(ix) = −i cosech(x)

sec(ix) = sech(x)

cot(ix) = −i coth(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cos(ix) = cosh(x)

tan(ix) = i tanh(x)

cosec(ix) = −i cosech(x)

sec(ix) = sech(x)

cot(ix) = −i coth(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cos(ix) = cosh(x)

tan(ix) = i tanh(x)

cosec(ix) = −i cosech(x)

sec(ix) = sech(x)

cot(ix) = −i coth(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cos(ix) = cosh(x)

tan(ix) = i tanh(x)

cosec(ix) = −i cosech(x)

sec(ix) = sech(x)

cot(ix) = −i coth(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Also we can replace x by ix in

sin(ix) = isinh(x), we get

sin i(ix) = isinh(ix)

⇒ sin(−x) = isinh(ix)

⇒ −sinx = isinh(ix)

⇒ i2sinx = isinh(ix)

∴ sinh(ix) = isinx

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Also we can replace x by ix in

sin(ix) = isinh(x), we get

sin i(ix) = isinh(ix)

⇒ sin(−x) = isinh(ix)

⇒ −sinx = isinh(ix)

⇒ i2sinx = isinh(ix)

∴ sinh(ix) = isinx

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Also we can replace x by ix in

sin(ix) = isinh(x), we get

sin i(ix) = isinh(ix)

⇒ sin(−x) = isinh(ix)

⇒ −sinx = isinh(ix)

⇒ i2sinx = isinh(ix)

∴ sinh(ix) = isinx

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Also we can replace x by ix in

sin(ix) = isinh(x), we get

sin i(ix) = isinh(ix)

⇒ sin(−x) = isinh(ix)

⇒ −sinx = isinh(ix)

⇒ i2sinx = isinh(ix)

∴ sinh(ix) = isinx

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Also we can replace x by ix in

sin(ix) = isinh(x), we get

sin i(ix) = isinh(ix)

⇒ sin(−x) = isinh(ix)

⇒ −sinx = isinh(ix)

⇒ i2sinx = isinh(ix)

∴ sinh(ix) = isinx

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Also we can replace x by ix in

sin(ix) = isinh(x), we get

sin i(ix) = isinh(ix)

⇒ sin(−x) = isinh(ix)

⇒ −sinx = isinh(ix)

⇒ i2sinx = isinh(ix)

∴ sinh(ix) = isinx

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Also we can replace x by ix in

sin(ix) = isinh(x), we get

sin i(ix) = isinh(ix)

⇒ sin(−x) = isinh(ix)

⇒ −sinx = isinh(ix)

⇒ i2sinx = isinh(ix)

∴ sinh(ix) = isinx

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cosh(ix) = cos(x)

tanh(ix) = i tan(x)

cosech(ix) = −i cosec(x)

sech(ix) = sec(x)

coth(ix) = −i cot(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cosh(ix) = cos(x)

tanh(ix) = i tan(x)

cosech(ix) = −i cosec(x)

sech(ix) = sec(x)

coth(ix) = −i cot(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cosh(ix) = cos(x)

tanh(ix) = i tan(x)

cosech(ix) = −i cosec(x)

sech(ix) = sec(x)

coth(ix) = −i cot(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cosh(ix) = cos(x)

tanh(ix) = i tan(x)

cosech(ix) = −i cosec(x)

sech(ix) = sec(x)

coth(ix) = −i cot(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cosh(ix) = cos(x)

tanh(ix) = i tan(x)

cosech(ix) = −i cosec(x)

sech(ix) = sec(x)

coth(ix) = −i cot(x)

N. B. Vyas Complex Numbers - 2

Relation between Circular & Hyperbolicfunctions

Similarly, we can prove

cosh(ix) = cos(x)

tanh(ix) = i tan(x)

cosech(ix) = −i cosec(x)

sech(ix) = sec(x)

coth(ix) = −i cot(x)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Identities of Hyperbolic functions can be derivedfrom the identities of circular functions byreplacing x by ix

Now sin2x+ cos2x = 1

sin2(ix) + cos2(ix) = 1⇒[i sinh(x)]2 + [cosh(x)]2 = 1

⇒ cosh2(x)− sinh2(x) = 1

Similarly we can obtain

sech2(x) = 1− tanh2(x)

cosech2(x) = coth2(x)− 1

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Identities of Hyperbolic functions can be derivedfrom the identities of circular functions byreplacing x by ix

Now sin2x+ cos2x = 1

sin2(ix) + cos2(ix) = 1⇒[i sinh(x)]2 + [cosh(x)]2 = 1

⇒ cosh2(x)− sinh2(x) = 1

Similarly we can obtain

sech2(x) = 1− tanh2(x)

cosech2(x) = coth2(x)− 1

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Identities of Hyperbolic functions can be derivedfrom the identities of circular functions byreplacing x by ix

Now sin2x+ cos2x = 1

sin2(ix) + cos2(ix) = 1⇒[i sinh(x)]2 + [cosh(x)]2 = 1

⇒ cosh2(x)− sinh2(x) = 1

Similarly we can obtain

sech2(x) = 1− tanh2(x)

cosech2(x) = coth2(x)− 1

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Identities of Hyperbolic functions can be derivedfrom the identities of circular functions byreplacing x by ix

Now sin2x+ cos2x = 1

sin2(ix) + cos2(ix) = 1⇒[i sinh(x)]2 + [cosh(x)]2 = 1

⇒ cosh2(x)− sinh2(x) = 1

Similarly we can obtain

sech2(x) = 1− tanh2(x)

cosech2(x) = coth2(x)− 1

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Identities of Hyperbolic functions can be derivedfrom the identities of circular functions byreplacing x by ix

Now sin2x+ cos2x = 1

sin2(ix) + cos2(ix) = 1⇒[i sinh(x)]2 + [cosh(x)]2 = 1

⇒ cosh2(x)− sinh2(x) = 1

Similarly we can obtain

sech2(x) = 1− tanh2(x)

cosech2(x) = coth2(x)− 1

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Identities of Hyperbolic functions can be derivedfrom the identities of circular functions byreplacing x by ix

Now sin2x+ cos2x = 1

sin2(ix) + cos2(ix) = 1⇒[i sinh(x)]2 + [cosh(x)]2 = 1

⇒ cosh2(x)− sinh2(x) = 1

Similarly we can obtain

sech2(x) = 1− tanh2(x)

cosech2(x) = coth2(x)− 1

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Identities of Hyperbolic functions can be derivedfrom the identities of circular functions byreplacing x by ix

Now sin2x+ cos2x = 1

sin2(ix) + cos2(ix) = 1⇒[i sinh(x)]2 + [cosh(x)]2 = 1

⇒ cosh2(x)− sinh2(x) = 1

Similarly we can obtain

sech2(x) = 1− tanh2(x)

cosech2(x) = coth2(x)− 1

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) =

2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) =

2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 =

1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) =

3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x)

, cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) =

4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =

3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =

2 tanh(x

2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

)

, cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =

1 + tanh2(x

2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)

tanh(x) =2tanh

(x2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)

N. B. Vyas Complex Numbers - 2

Hyperbolic Identities

Also sin2x = 2 sinx cosx

by replacing x by ix

we get sinh(2x) = 2 sinh(x) cosh(x)

Similarlycosh(2x) = cosh2(x) + sinh2(x) = 2cosh2(x)− 1 = 1 + 2sinh2(x)

sinh(3x) = 3sinh(x)+4sinh3(x) , cosh(3x) = 4cosh3(x)−3cosh(x)

tanh(3x) =3tanh(x) + tanh3(x)

1 + 3tanh2(x)

sinh(x) =2 tanh

(x2

)1− tanh2

(x2

) , cosh(x) =1 + tanh2

(x2

)1− tanh2

(x2

)tanh(x) =

2tanh(x

2

)1 + tanh2

(x2

)N. B. Vyas Complex Numbers - 2

Logarithm of a Complex Number

If z = ew, then we write w = lnz, called the natural logarithm ofz. Thus the natural logarithmic function is the inverse of theexponential function and can be defined byw = lnz = ln(rei(θ+2kπ)) = lnr + i(θ + 2kπ)

∀zε{, logz = ln|z|+ iarg(z)

N. B. Vyas Complex Numbers - 2