The value of $\frac{(-1+i\sqrt 3)^{15}}{(1-i)^{20}}+\frac{(-1-i\sqrt 3)^{15}}{(1+i)^{20}}$ is

Question

The expression can be written as

$$\frac{(2w)^{15}}{(\cos\frac{\pi}{4}-i\sin\frac{\pi}{4})^{20}}+\frac{(2w^2)^{15}}{(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})^{20}}$$

Where w is a cube root of $1$ $$=2^{15}\left[\frac{w^{15}}{\cos 5\pi - i\sin 5\pi}+\frac{w^{30}}{\cos 5\pi + i\sin 5}\right ]$$

$$=2^{15}[-1-1]$$ $$=-2^{16}$$

The answer given is $-64$. What is going wrong?

Answer 1

You can make your calculations a bit easier by realizing that the expression is $z + \bar z = 2\Re{(z)}$ with $z = \frac{(-1+i\sqrt 3)^{15}}{(1-i)^{20}}$.

Now, using

• $-1+i\sqrt{3} = 2e^{\frac{2}{3}\pi i}$
• $1-i = \sqrt{2}e^{-\frac{1}{4}\pi i}$

you get

$$2\Re{(z)} = 2\cdot \Re \left( \frac{2^{15}}{2^{10}e^{-\pi i}}\right) = -2^6 = -64$$

Answer 2

It is $$(-1+i\sqrt{3})^{15}=32768$$ and $$(1-i)^{20}=-1024$$