solve z^{7} =1
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.
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