Calculate $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$.
Here is what I did so far:
I'm trying to transform $z$ into its trigonometric form, so I can use De Moivre's formula for calculating $z^{2016}$.
Let $z = \frac{-1 + i\sqrt 3}{1 + i}$. This can be rewritten as $\frac{\sqrt 3 – 1}{2} + i\frac{\sqrt 3 + 1}{2}$.
$$z = r(\cos \phi + i \sin \phi)$$
$$r = |z| = \sqrt 2$$
$$\phi = \arctan {\sqrt 3 + 1}$$
Now, I don't know what to do with that $\sqrt 3 + 1$. How do I calculate $\phi$ ?
Thank you in advance!
Best Answer
Here is another approach: Why don't you "distribute" that exponent on the numerator and denominator? Then raise both numerator and denominator to the power 2016. The thing is that both $\arctan(-\sqrt{3})$ as well as $\arctan 1$ are well known angles. From there you can apply your DeMoivre. Once you have those new numerators and denominators, you can simply divide. I will do the denominator for you: $r=\sqrt{2}$ and $\theta=45°$, so to the power 2016 is $2^{1008}(\cos(2016(45°))+i\sin(2016(45°)))$ which is $2^{1008}(1+0i)$. Can you do the numerator?