## Loser66 one year ago Find real and imaginary part of $$(\dfrac{-1-i\sqrt3}{2})^6$$ by polar expression. Please, help

1. Loser66

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

2. IrishBoy123

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

3. Loser66

Yes|dw:1441922670446:dw|

4. anonymous
5. anonymous

idk why i typed this :O

6. anonymous

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

7. Loser66

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

8. Loser66

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?

9. anonymous

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

10. IrishBoy123

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

11. IrishBoy123

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

12. IrishBoy123

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