Complex Numbers and Euler’s Identitybanach.millersville.edu/~bob/math171/Complex/main.pdfComplex...

Complex Numbers and Euler’s IdentityMATH 171 Freshman Seminar for Mathematics Majors

J. Robert Buchanan

Department of Mathematics

2010

J. Robert Buchanan Complex Numbers and Euler’s Identity

Background

Easy: solve the equation 0 = 1 − z2.

0 = 1 − z2

= (1 − z)(1 + z)

1 = z or − 1 = z


Background

Easy: solve the equation 0 = 1 − z2.

0 = 1 − z2

= (1 − z)(1 + z)

1 = z or − 1 = z

Not (as) easy: solve the equation 0 = 1 + z2.

0 = 1 + z2

= −1 − z2

= (√−1 − z)(−

√−1 + z)√

−1 = z or −√−1 = z

Define i =√−1, then z = ±i .


Complex Numbers

Definition

A number of the form z = a + bi where a and b are realnumbers and i =

√−1 is called a complex number . If a = 0

and b 6= 0 so that z = bi , then z is called an imaginarynumber .


Picturing Complex Numbers

z = a + bi

Ha,bL

x

i y


Polar Representation (1 of 2)

z = reiθ

Ha,bL

Θ

r

x

i y


Polar Representation (2 of 2)

z = a + bi = reiθ

where

θ = tan−1(

ba

)

and

r =

{√

a2 + b2 if a > 0,−√

a2 + b2 if a < 0.


Examples (1 of 3)

Example

Find the polar representation of the following complex numbers.

−1 + i

2 − i

−3 − i


Examples (2 of 3)

H-1,1L

H2,-1LH-3,-1L

x

i y


Examples (3 of 3)

Example

Find the polar representation of the following complex numbers.

−1 + i = −√

2e−iπ/4

2 − i =√

5ei tan−1(−1/2) ≈√

5e−0.463648i

−3 − i = −√

10ei tan−1(1/3) ≈ −√

10e0.321751i


Complex Arithmetic

If z1 = a + bi and z2 = c + di , then

z1 + z2 = (a + c) + (b + d)i

z1 − z2 = (a − c) + (b − d)i

z1z2 = (ac − bd) + (ad + bc)iz1

z2=

(ac + bd) + (bc − ad)ic2 + d2 ,

provided c2 + d2 6= 0.


Complex Addition (1 of 3)

(−3 + 4i) + (5 − 2i) = 2 + 2i

z1

z2

H-3,4L

H5,-2L

x

i y



(−3 + 4i) + (5 − 2i) = 2 + 2i

z1

z2

z1+z2

H-3,4L

H2,2L

H5,-2L

x

i y



(−3 + 4i) + (5 − 2i) = 2 + 2i

z1

z2

z1+z2

H-3,4L

H2,2L

H5,-2L

z2

z1

x

i y


Complex Multiplication

z1z2 = r1eiθ1r2eiθ2 = (r1r2)ei(θ1+θ2)

Θ1

r1Θ2r2r1r2

Θ1+Θ2

x

i y


Challenge

Represent z = −1 = −1 + 0i in polar form.


Challenge


Since r =√

(−1)2 + 02 = 1, then

−1 = eiπ.


Challenge


Since r =√

(−1)2 + 02 = 1, then

−1 = eiπ.

Rearranging the equation above yields an equation relating fiveof the most important constants in mathematics.

eiπ + 1 = 0


Euler’s Identity

eiθ = cos θ + i sin θ

Θ

r

r ei Θ=rHcosHΘL+ i sinHΘLL

x

i y


Commemorative Stamp


Complex Exponentiation

Use Euler’s Identity (eiθ = cos θ + i sin θ) to

express z = i in polar form, and


Complex Exponentiation

Use Euler’s Identity (eiθ = cos θ + i sin θ) to

express z = i in polar form, and

evaluate i i .


Infinite Series

In Calculus II you will learn to express the function ex as theinfinite series :

ex = 1 +x1!

+x2

2!+

x3

3!+ · · ·

where n! = (1)(2)(3) · · · (n).


Infinite Series


ex = 1 +x1!

+x2

2!+

x3

3!+ · · ·

where n! = (1)(2)(3) · · · (n).

This infinite series holds for real and complex values of x .


Infinite Series


ex = 1 +x1!

+x2

2!+

x3

3!+ · · ·

where n! = (1)(2)(3) · · · (n).

This infinite series holds for real and complex values of x .

For example,

−1 = eiπ = 1 +iπ1!

+(iπ)2

2!+

(iπ)3

3!+ · · ·


Expressing eiπ as a Series

−1 = 1 +iπ1!

+(iπ)2

2!+

(iπ)3

3!+

(iπ)4

4!+

(iπ)5

5!+ · · ·

= 1 + iπ − π2

2− i

π3

6+

π4

24+ i

π5

120− · · ·


Expressing eiπ as a Series

−1 = 1 +iπ1!

+(iπ)2

2!+

(iπ)3

3!+

(iπ)4

4!+

(iπ)5

5!+ · · ·

= 1 + iπ − π2

2− i

π3

6+

π4

24+ i

π5

120− · · ·

Note: the series consists of alternating real and purelyimaginary terms.


Expressing the Series for eiπ as a Graph

−1 = 1 + iπ − π2

2− i

π3

6+

π4

24+ i

π5

120− · · ·

-4 -3 -2 -1 1x

-2

-1

1

2

3

i y


Quadratic Map

Suppose f (z) = z2 + c where z and c can be complexnumbers.Similar to the Newton’s Method formula we may iterate thequadratic function f (z).Starting with z0 we define

zn = f (zn−1) = z2n−1 + c

for n = 1, 2, . . ..


Quadratic Map

Suppose f (z) = z2 + c where z and c can be complexnumbers.Similar to the Newton’s Method formula we may iterate thequadratic function f (z).Starting with z0 we define

zn = f (zn−1) = z2n−1 + c

for n = 1, 2, . . ..

For example, if z0 = 0 and c = 12 i then

z1 =12

i

z2 = −14

+12

i

z3 = − 316

+14

i

...


Quadratic Map (Graph)

-0.25 -0.20 -0.15 -0.10 -0.05x

0.1

0.2

0.3

0.4

0.5

i y


General Exponentiation

Suppose x and y are two real numbers and suppose y > x > 0.

Question: which is larger xy or yx ?


General Exponentiation

Suppose x and y are two real numbers and suppose y > x > 0.

Question: which is larger xy or yx ?

Remember one of our principles of mathematical inquiry, trysome examples in order to gain insight into a complicatedquestion.

Let x = 2 and try y = 3, y = 4, and y = 5.


Equality (1 of 3)

Since for some choices of 0 < x < y ,

xy > yx

while for othersxy < yx

we may be curious about when xy = yx .


Equality (2 of 3)

xy = yx

ln(xy ) = ln(yx )

y ln x = x ln y

Since y > x > 0 then y = kx where k > 1. Substitute this intothe last equation above and solve for x and y in terms of k .


Equality (3 of 3)

For k > 1,

x = k1/(k−1)

y = kk/(k−1).

2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7

2.8

3.0

3.2

3.4

3.6

3.8

4.0

x

y


Limits as k → 1+

Use l’Hôpital’s Rule to find

limk→1+

x = limk→1+

k1/(k−1)

limk→1+

y = limk→1+

kk/(k−1).


Limits as k → 1+

Use l’Hôpital’s Rule to find

limk→1+

x = limk→1+

k1/(k−1)

limk→1+

y = limk→1+

kk/(k−1).

limk→1+

k1/(k−1) = e

limk→1+

kk/(k−1) = e


Summary

x=yxy=yx

He,eL

HemptyL y<x

yx>xy

xy>yx

0 1 2 3 4 5 6

0

1

2

3

4

5

6

x

y


Students and Complex Numbers

Student c Student cBacchi −11

8 + i j8 Bongiovanni −5

4 + i j8

Cilladi −98 + i j

8 Cox −1 + i j8

Crider −78 + i j

8 de Kok −34 + i j

8Hansford −5

8 + i j8 Hild −1

2 + i j8

Junkin −38 + i j

8 Keglovits −14 + i j

8Kibler −1

8 + i j8 Konowal 1

8 + i j8

Leber 14 + i j

8 Longo 38 + i j

8Mecutchen 1

2 + i j8 Miller, B 5

8 + i j8

Miller, S 34 + i j

8 Nguyen 78 + i j

8Reed 1 + i j

8 Smeltz 98 + i j

8Starner 5

4 + i j8 Visek 11

8 + i j8

Williard 32 + i j

8 Zipko 138 + i j

8


Results

Name:

c =

Results:

1 2 3 4 5 6 7 8 9 10 11 12

13 14 15 16 17 18 19 20 21 22 23 24


Homework

Referring to the polygonal spiral approaching −1 on slide28, find the total length of the spiral.

For the complex number c you have been assigned andstarting with z0 = 0 iterate the quadratic map f (z) = z2 + cten times (or less if r =

√a2 + b2 > 2) for j = 1, 2, . . . , 24.

If all the iterates of the quadratic map have a magnitude ofless than 2 record a result of 1, else record a result of 0.


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