008 Complex Numbers

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    Complex NumbersComplex Numbers

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    Consider the quadratic equation x 2 + 1= 0.

    Solving for x , gives x 2 = 112 ! x

    1! x

    Complex Numbers

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    Since there is not real number whoseSince there is not real number whosesquare issquare is - -1 , the equation has no1 , the equation has noreal solution. French mathematicianreal solution. French mathematician

    Rene Descartes (1596Rene Descartes (1596 - -1650)1650)proposed thatproposed that i i be defined suchbe defined suchthat ,that ,

    1!i

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    1!i

    Complex Numbers

    12 !i Note that squaring both sides yields:therefore

    and

    so

    and

    iiiii !!! *1* 132

    1)1(*)1(* 224 !!! iii

    iiiii !!! *1*45

    1*1* 2246 !!! iiii

    A nd so on

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    Finding Powers of Finding Powers of i i

    T he successive powers of i rotate through the four values of i , -1, - i , and 1.

    i n = i if

    n = 1, 5 , 9 ,

    i n = - 1 if

    n = 2, 6 , 10,

    i n = - i if

    n = 3 , 7 , 11,

    i n = 1 if n

    = 4 , 8 , 12,

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    then,1-I !i

    12 !i

    ii !3

    14 !i

    ii !5

    16

    !i

    ii !7

    18 !i

    * For larger exponents , divide the exponent by

    4 , then use theremainder as your

    exponent instead.

    Example: ?23 !i

    3oremainder aith5423

    !

    .etcii -hichuseSo, 3 !

    ii !23

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    R eal NumbersImaginary Numbers

    Real numbers and imaginarynumbers are subsets of the set of complex numbers.

    Complex Numbers

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    Imaginary UnitImaginary UnitUntil now , you have always been toldthat you cant take the square root of

    a negative number.If you useimaginary units , you can!

    The imaginary unit is .=I t is used to write the square root of a negative number.

    1

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    Property of the square rootProperty of the square rootof negative numbersof negative numbers

    I f r is a positive real number , then

    r ir ! r i

    Examples:

    !3 3i !4 !4i i2

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    ExamplesExamples2

    )3( 1. i22 )3(i!

    )3*3(1!

    )3(1!3!

    26103 olve 2.2

    ! x363 2 ! x

    122 ! x

    122 ! x12i x s!

    32i x s!

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    Complex NumbersComplex NumbersA complex number has a real part & an imaginary part.

    Standard form is: b ia

    Real part Imaginary part

    Example: 5+4iExample: 5+4i

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    Adding and SubtractingAdding and Subtracting

    To Add or Subt rac t C omplex Nu m b ers

    1. Change all imaginary numbers to b i form .

    2. A dd (or subtract ) the real parts of the complexnumbers .

    3. A dd (or subtract ) the imaginary parts of the

    complex numbers .4. Write the answer in the form a + b i .

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    Adding and SubtractingAdding and Subtracting(add or subtract the real parts , then(add or subtract the real parts , thenadd or subtract the imaginary parts)add or subtract the imaginary parts)

    Ex: )33()21( ii)32()31( ii!

    i52!

    Ex: )73()32( ii)73()32( ii!

    i41!

    Ex: )32()3(2 iii)32()23( iii!

    i21!

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    MultiplyingMultiplying

    To Mu lti ply C omplex Nu m b ers

    1. Change all imaginary numbers to b i form .

    2. Multiply the complex numbers as you wouldmultiply polynomials .

    3. Substitute 1 for each i 2.

    4. Combine the real parts and the imaginary parts . Write the answer in a + b i form .

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    MultiplyingMultiplyingTreat the is like variables , thenTreat the is like variables , then

    change any that are not to thechange any that are not to thefirst powerfirst power

    Ex: )3( ii2

    3 ii!)1(3! i

    i31!

    Ex: )26)(32( ii2618412 iii!

    )1(62212! i

    62212! ii226!

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    CAUTIO N!CAUTIO N!

    ?24 !

    !! 2424 ii

    !22 2i 22

    824 {

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    DividingDividing

    To Divid e C omplex Nu m b ers

    1. Change all imaginary numbers to b i form .

    2. R ationalize the denominator by multiplying boththe numerator and the denominator by theconjugate of the denominator .

    3. Substitute 1 for each i 2.

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    DividingDividing

    E xamples:

    !i

    i

    3434

    !i

    i

    i

    i

    3434

    3434

    !2316

    3434i

    ii

    !916

    924162

    ii25

    247 i

    !

    53

    5!

    53

    5i

    !

    53

    53

    53

    5i

    i

    i

    !259

    )53(5i

    i!

    59)5515 i

    14)5515 i

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    i

    i

    i

    i E x

    2121

    *21

    113 :

    )21)(21()21)(113(

    ii

    ii!

    2

    2

    4221221163iii

    iii!

    )1(41

    )1(2253!

    i

    412253

    !i

    5

    525 i!

    55

    525 i

    !

    i! 5

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    EVALUATE theFOLLOWING

    8

    15

    27

    52

    12

    1.

    2.

    3. 2

    4.5.

    i

    i

    i

    i

    i

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    C hange to biform 1. 36

    2. 120

    3. 48

    4. 6005. 4 64

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    1 . ( 2 8 ) (3 5 )

    2 . (5 8 ) ( 4 2 ) (3 )

    3 . 3 ( 2 5 )

    4 . ( 4 3 )(2 5 )

    45 .

    23

    6 .

    i i

    i i i

    i i

    i i

    i

    ii

    i

    P erform the followingoperations