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Thread: 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: $\displaystyle z= r cis(\theta)$ for "cos+ i s sin") which can also be written as $\displaystyle re^{i\theta}$ 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: $\displaystyle [r(cos(\theta)+ i sin(\theta))]^n= r^n (cos(n\theta)+ i sin(n\theta)$ or, in exponential form, $\displaystyle (re^{i\theta}= r^n e^{i n\theta})$.

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

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

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

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

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

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