Showing the Summation of $(\frac{w}{2})^k$ where w is a complex root

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I got the correct answer for (i) and (ii) and the problem is with third part. I cant find my mistake.

Since the third part is related to the second part, I will mention its answer.

The roots are :

$$z=2e^{(-\frac{2i\pi}{15}+\frac{2ki\pi}{5})}$$ where $k=0,1,2,3,4$


Now here is My Attempt to (iii):

$$w=2e^{(-\frac{2i\pi}{15}+\frac{2k_0i\pi}{5})}$$

The Summation follows the sum of a G.P, so therefore applying the formula:

$$S_5=1\frac{1-(\frac{w}{2}^5)}{1-\frac{w}{2}}$$

$$\frac{\frac{32-w^5}{32}}{\frac{2-w}{2}}$$

$$\frac{32-w^5}{16(2-w)}=\frac{32-32e^{\frac{2i\pi}{3}+2k_1\pi}}{16(2-w)}$$

$$\frac{32-32e^{\frac{2i\pi}{3}}}{16(2-w)}=\frac{2-2(\frac{-1}{2}+\frac{\sqrt{3}i}{2})}{2-w}=\frac{3-i\sqrt3}{2-w}$$


As you can see I get - instead of +. Where did I go wrong? Please help.

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When replacing $$w=2e^{(-\frac{2i\pi}{15}+\frac{2k_0i\pi}{5})}$$ in $w^5$ you use $$w^5=32e^{\frac{2i\pi}{3}+2k_1\pi}$$ instead of the correct version $$w^5=32e^{-\frac{2i\pi}{3}+2k_0\pi}.$$

If you retrace your steps with this corrected value, everything should be fine.