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Math Help - simple complex number

  1. #1
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    simple complex number

    if z=x+yi, find all possible values of z such that z^3=1

    please enlighten me, no idea where to start.
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  2. #2
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    Quote Originally Posted by z1llch View Post
    if z=x+yi, find all possible values of z such that z^3=1

    please enlighten me, no idea where to start.
    It is simplest to write z in "polar form"- [/tex]z= r(cos(\theta)+ i sin(\theta))[/tex] (Engineering notation: z= r cis(\theta) for "cos+ i s sin") which can also be written as re^{i\theta} where "r" is the straight line distance from 0 to x+ iy in the "complex plane" ((0,0) to (x,y)) and [tex]\theta[/itex] is the angle that line makes with the positive real axis (positive x-axis).

    Then you can use De Moivres theorem: [r(cos(\theta)+ i sin(\theta))]^n= r^n (cos(n\theta)+ i sin(n\theta) or, in exponential form, (re^{i\theta}= r^n e^{i n\theta}).

    There, n can be any real number including 1/3: z^{1/3}= r^{1/3}(cos(\theta/3)+ i sin(\theta/3)).

    Of course, the solutions to z^3= 1 are z= (1)^{1/3}.

    "1" is ON the real-axis to \theta= 0 and its distance from 0 is 1 so r= 1: z^3= 1= 1(cos(0)+ i sin(0)).

    That's obviously true: cos(0)= 1, sin(0)= 0. It is also obvious that z= 1^{1/3}(cos(0/3)+ i sin(0/3)= 1(cos(0)+ i sin(0))= 1. Of course you knew that one solution of z^3= 1 was z= 1.

    Now the good part: Since sine and cosine are periodic with period 2\pi, z^3 can also be written as 1(cos(2\pi)+ i sin(2\pi)) or as 1(cos(4\pi)+ i sin(4\pi)) but dividing by 3, so that the sine and cosine are of \frac{2\pi}{3} and \frac{4\pi}{3} gives different results! That's how you can get the two other roots.

    (Adding another 2\pi would give 6\pi and dividing that by 3 gives 2\pi again and so not a new solution. If you keep adding " 2\pi", you cycle though those same three solutions.)
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