Find all $z$ such that $|\cos z|^2+|\sin z|^2=4$

Question

I need to solve for $z$ with $|\cos z|^2+|\sin z|^2=4$ I know $\cos z =\frac{1}{2}(e^{iz}+e^{-iz})$ and $\sin z = \frac{1}{2i}(e^{iz}-e^{-iz})$ but I'm not sure if this is helpful because I don't know how to split it into Re$(z)$ and Im$(z)$ to find $|\text{cos}\ z|$ and $|\text{sin}\ z|$

Answer 1

Since\begin{align}|\cos z|^2+|\sin z|^2&=\cos(z)\overline{\cos(z)}+\sin(z)\overline{\sin(z)}\\&=\cos(z)\cos\left(\overline z\right)+\sin(z)\sin\left(\overline z\right)\\&=\cos\left(z-\overline z\right)\\&=\cos\bigl(2\operatorname{Im}(z)i\bigr)\\&=\cosh(2\operatorname{Im}z),\end{align}all you have to do is to solve the equation $\cosh(2\operatorname{Im}z)=4$. Can you take it from here?

Answer 2

Let $z=x+iy$. Then,

$$|\cos z|^2 = \cos^2x\cosh^2y+\sin^2 x \sinh^2 y$$ $$|\sin z|^2 = \sin^2x\cosh^2y+\cos^2 x \sinh^2 y$$

and

$$|\cos z|^2 + |\sin z|^2 =\cosh^2 y + \sinh^2 y = \cosh 2y =4$$

Thus, the solutions are

$$z=x+\frac i 2\cosh^{-1}4$$