solve z

^{7}=1-
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solve z^{7} =1

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According to the fundamental theorem of algebra, the equation z^{7}=1 has 7 roots, called the roots of unity. Only one of them is obvious: z=1. The other six require polar form. First write 1 in polar form:

1 = e^{Pi*in}, where n is any integer.

Then set

z^{7}=e^{Pi*in}

and take the seventh root:

z=e^{Pi*in/7}

Now substitute in values for n. You only need the values n=0,1,2,3,4,5, and 6, because the answer will repeat itself for all other n (due to the periodicity of complex exponentials).

For n=0, you get the obvious solution,

z_{0}=e^{0}=1.

For n=1, you get

z_{1}=e^{Pi*i/7},

which you can write in standard form using Euler's identity:

z_{1}=cos(Pi/7)+i sin(Pi/7)

In this way, you get the seven seventh roots of unity.

If you graph them in the complex plane, you will find they all lie equally spaced on the unit circle.

sorry, didn't realize it was z^{7}. only saw z7 in the list of questions.

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