What I am missing here?

Question

I was solving a problem about complex numbers. After trying for sometime, I looked at the solution. I didn't understand the logic behind their reasoning. Those expressions don't look equal at least to me.

What I am missing? Is it about the fact that the argument of a complex number can be anything between -π and π.

Answer 1

$e^{i \frac{125\pi}{12}} = e^{i \frac{5\pi}{12}}$

This follows because $\frac{125\pi}{12} = \frac{5\pi}{12} + 10 \cdot \pi = \frac{5\pi}{12} + 5 \cdot 2\pi$, and $e^{2\pi i} = 1$.

$e^{i \frac{5\pi}{12}} = e^{i \left(\frac{\pi}{2}-\frac{\pi}{12}\right)}$

$\frac{5}{12}=\frac{1}{2}-\frac{1}{12}$

$e^{i \left(\frac{\pi}{2}-\frac{\pi}{12}\right)} = ie^{-i\frac{\pi}{12}}$

This is because $e^{i \frac{\pi}{2}} = \cos\frac{\pi}{2}+i\sin\frac{\pi}{2}= i$.

$ie^{-i\frac{\pi}{12}}=ie^{i \left(\frac{\pi}{4}-\frac{\pi}{3}\right)}$

$-\frac{1}{12}=\frac{1}{4}-\frac{1}{3}$

$ie^{i \left(\frac{\pi}{4}-\frac{\pi}{3}\right)} =ie^{i \frac{\pi}{4}}e^{-i \frac{\pi}{3}}$

Property of exponential function.

You could continue to simplify from here by noting that $$e^{i \frac{\pi}{4}} = \cos\frac{\pi}{4}+i\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$$ and $$e^{-i \frac{\pi}{3}} = \cos -\frac{\pi}{3} + i \sin -\frac{\pi}{3}=\frac{1}{2}-i\frac{\sqrt{3}}{2}.$$ So we get \begin{align*} e^{i \frac{125\pi}{12}} &= i\left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\\ &=\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{2}+\sqrt{6}}{4}i. \end{align*}