Determine all complex numbers satisfying the equation

Question

The equation:
$$e^{iz} = 4\cos(z) + 3$$ This is what I have tried so far: $$e^{iz} = 4\left(\frac{e^{iz}+e^{-iz}}{2}\right)+3$$ $$e^{iz} = 2e^{iz}+2^{-iz}+3$$ $$e^{2iz} = 2e^{2iz}+2+3e^{iz}$$ $$e^{2iz}+3e^{iz}+2=0$$ Now, I substitute $e^{iz}$ with $x$, resulting in $x=-1$ or $x=-2$. This leaves me at $e^{iz}=-1$ or $e^{iz}=-2$
I'm not sure how to continue from here. I can get rid of the $e$ by taking the natural logarithm, but that makes further calculations very difficult.

Answer 1

We use the following well-known properties of the complex exponential function: $\forall z \in \mathbb{C}, \forall z' \in \mathbb{C}:$

• $e^{z} e^{z'}=e^{z+z'}$;
• $e^z=1\iff\exists k\in \mathbb{Z}, z=2ik\pi$.

From these properties, it follows that : $\forall z \in \mathbb{C}, e^z\neq0$; $\forall z \in \mathbb{C}, \forall z' \in \mathbb{C}, e^z=e^{z'}\iff \exists k\in \mathbb{Z},z=z'+2ik\pi$.

So, $e^{iz}=-1=e^{i\pi}\iff \exists k\in \mathbb{Z},iz=i\pi+2ik\pi\iff \exists k\in \mathbb{Z},z=\pi+2k\pi$ $e^{iz}=-2=e^{i\pi}e^{log2}=e^{i\pi+log2}\iff \exists k\in \mathbb{Z},iz=i\pi+log2+2ik\pi\iff \exists k\in \mathbb{Z},z=\pi+2k\pi-ilog2$