Find $a,b$ from $\sin(2a)=\sin(2b)$, $a^2+b^2=1$

Question

So I have these complex numbers $z$ with the modulus $1$.

$$\sin(z+\overline{z})-\cos(\frac{\pi}{2}+i(z-\overline{z}))=0$$ I need to find the element $Re^{4}z+Im^{4}z$

I started with $z=a+bi$ and I got $\sin(2a)=\sin(2b)$.Now I get stuck when I need to find a and b.I mean I know that $a^2+b^2=1$ and I can see the $a=b=-+\frac{1}{\sqrt2}$ but I want to know the algorithm.

How to continue?

Answer 1

By trigonometry one has $\sin 2a = \sin 2b$ if and only if either $2a = 2b + 2n\pi$ for some integer $n$ or $2a + 2b = \pi + 2n\pi$ for some integer $n$. The only way these are compatible with $a^2 + b^2 = 1$ is if $a = b = \pm {\sqrt{2} \over 2}$.

Answer 2

Hint: Use that $$\sin(2a)-\sin(2b)=2 \sin (a-b) \cos (a+b)$$