find all the solutions of $ e^{a+ib}=e^{a-ib}+2i $ $Re(a+ib) \le 0 $

Question

find all the solutions in the complex field of the system

$$ e^{a+ib}=e^{a-ib}+2i $$ $$Re(a+ib) \le 0 $$

where $a+ib$=z and a-ib=conj(z)

Answer 1

I will try to explain this with as much detail as possible, perhaps overwhelmingly so. We have: $$ e^{a+ib}=e^{a-ib}+2i $$ $$Re(a+ib) \le 0 $$

We first try to add a condition to the two given by looking at the function $f(z) = e^z-e^\bar z $ where $\bar z$ denotes complex conjugation. If $z= a+ib$ we find a more insightful form for $f$: $$e^{a+ib} -e^{a-ib}\\ e^a(\cos b + i\sin b)-e^a(\cos (-b) +i\sin (-b))\\ e^a(\cos b + i\sin b)-e^a(\cos b -i\sin b)\\ 2ie^a\sin b$$

So if $$2ie^a\sin b = 2i$$ then $$e^a\sin b =1$$

We know that $\sin \theta \leq 1 $ for all $\theta$, which implies $e^a$ must be greater than or equal to $1$ for equality to hold. If $a<0$ this is impossible, because $e^x < 1$ for $x$ real and negative. This implies $a=0$ and we only have to look at $e^{ib}$ for $ b \in \Bbb{R}$.

If you then think of conjugation as reflection across the real line on the complex plane, we know that $e^z$ must lie in the line $y=1$ for the first condition to hold. The only point on this line and on the unit circle (that is, the points $e^{ib}$ with real $b$), is $i$. We therefore have the equation: $$e^{ib} = i$$

Which has solutions $$b = {\pi \over 2} + 2\pi n, \quad n \in \Bbb{Z}$$

Making your final set of solutions $$z = i({\pi \over 2} + 2\pi n), \quad n \in \Bbb{Z}$$

Answer 2

$$\begin{align*} e^{a+ib}=e^{a-ib}+2i\\Re(a+ib) \leq 0\end{align*} $$

But

$$e^{a+ib}=e^{a-ib}+2i\Longleftrightarrow e^a\sin b=1 $$

so any pair of the form

$$(a,b)=\left(0\,,\,\frac{(4n+1)\pi}{2}\right)\,\,,\,n\in\Bbb Z$$

is a solution, and since with $\,a<0\,$ there can't be solution (why?), the above are all the solutions.