Find roots of $f = e^{2z}-1,$ where $z = x+iy$

Question

Calculate the roots of $f = e^{2z}-1$, where $z = x+iy$.

If I use $e^{2z}-1=0$, then I get $2z = \ln(1)$ so $x = iy$, but is this correct?

Answer 1

Observe that

$$e^{2z}=e^{2x}e^{2iy}=1\implies e^{2x}=1 \implies x=0.$$

So we have $$e^{2iy}=1\implies 2y=2n\pi, \quad n\in \mathbb{Z}.$$

So $$z=in\pi.$$

Answer 2

Observe that (complete detalis)

$$e^z=e^{x+iy}=e^x(\cos y+i\sin y)=1\iff\begin{cases}e^x\sin y=0\iff &y=k\pi,\,k\in\Bbb Z\\{}\\ e^x\cos y=1\end{cases}$$

and knowing that it must be $\;y=k\pi\;,\;\;k\in\Bbb Z\;$ from the first line above, it follows that

$$e^x\cos k\pi=1\iff x=0,\,\,k\;\,\text{is even}$$

Well, now try to attack your problem again taking the above into account.

Answer 3

$\ln$ is a "multi valued" function. In particular, $ln(1)"="\{2\pi i n|n\in\mathbb{Z}\}$. I use $"="$ since the precise interpretation of this statement depends on the context and your definition. But you can think about it as follows:

$$\forall n\in\mathbb{Z}:e^{2\pi i n}=1$$ So if we denote $$f:\mathbb{C}\to \mathbb{C}:x\mapsto e^x$$ then $f$ is not injective, and hence does not have a well defined inverse function. However, choosing a pre image, for example $1$, and a "cut", for example the negative $x$-axis, specifies a "branch". This means that this data gives a function $$ln_{1,(-\infty,0]}:\mathbb{C}\setminus (-\infty,0]\to \mathbb{C}$$ such that $$e^{ln_{1,(-\infty,0]}(x)}=x$$ However, there are infinitely many functions that satisfy this property, corresponding to different choices of the pre image and cut. There is an arguably more elegant solution to this problem in the realm of complex geometry and anlysis, namely the notion of a Riemann surface, but that's too long a story to tell here.

So, coming back to your question, your mistake is that it is not true that $$e^z=1\Rightarrow z=0$$