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ra1l [238]
3 years ago
8

If

"absmiddle" class="latex-formula"> and z^{6}=a+bi, then a = _ and b = _
Mathematics
1 answer:
Luden [163]3 years ago
3 0

Rewriting z in polar form makes this trivial.

z=|z|e^{i\mathrm{arg}(z)}

We have

|z|=\sqrt{(-1)^2+(-\sqrt3)^2}=2

\mathrm{arg}(z)=\tan^{-1}(-1,-\sqrt3)=-\dfrac{2\pi}3

(not to be confused with the standard inverse tangent function \tan^{-1}x. Here \tan^{-1}(x,y) is the inverse tangent function that takes into account position in the coordinate plane; look up "atan2" for more information)

So we have

z=-1-\sqrt3\,i=2e^{-2\pi/3\,i}

Then

z^6=2^6\left(e^{-2\pi/3\,i}\right)^6=64e^{-4\pi\,i}=64

so that a=64 and b=0.

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