Find real and imaginary part of $(\dfrac{-1-i\sqrt3}{2})^6$ by polar expression.
Please, help

Question

Find real and imaginary part of $(\dfrac{-1-i\sqrt3}{2})^6$ by polar expression.
Please, help

loser66 · Answer

oh, I know what is wrong, it should be $z = |-1| e^{i4\pi/3}$ right?

irishboy123 · Answer

$(\dfrac{-1-i\sqrt3}{2})^6$

loser66 · Answer

Yes|dw:1441922670446:dw|

anonymous · Answer

anonymous · Answer

idk why i typed this :O

anonymous · Answer

and @Loser66  the polar form  is the one with cosine and sine not exponential Form

loser66 · Answer

But we can find the real part and imaginary part by polar, right?

loser66 · Answer

As above, I have $z = e^{4i\pi/3}$ , then $z^6 = e^{i8\pi}= cos (8\pi) + isin(8\pi)$ 
Hence R(z) =1 and Im (z) =0, right?

anonymous · Answer

yeah sure 
$ (x+iy)^n=(r\cos n \theta+i r \sin n \theta) $

irishboy123 · Answer

$$(\dfrac{-1-i\sqrt3}{2})^6$$
$$= (-\frac{1}{2} -i\frac{\sqrt{3}}{2})^6$$
$$=(e^{i \frac{4 \pi}{3}})^6$$
$$=e^{i 8 \pi} = 1$$

irishboy123 · Answer

find the 6 sixth roots of 1, next

that's more frustrating.

irishboy123 · Answer

|dw:1441924027440:dw|