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

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)

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Hint: If $e^{a+ib}=x+yi$ then $e^{a-ib}=x-yi$. What can $y$ be? – Thomas Andrews Jan 23 '13 at 14:49
Have you tried anything here? What does $\mathrm{Re}(a+ib)=a\leq 0$ mean to the values $e^{a+ib}$ and $e^{a-ib}$ – Thomas Andrews Jan 23 '13 at 14:56
Or you can write $e^{a+bi} = e^a (cosb + isinb)$ and see what happens... – Alan Simonin Jan 23 '13 at 14:56
Since the reason for the bounty is "The current answers do not contain enough detail", it would be appropriate to say what exactly it is that they are lacking. – Antonio Vargas Jan 25 '13 at 22:35
Exactly @AntonioVargas: I'd like to know what "detail" is lacking. – DonAntonio Jan 25 '13 at 22:48

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 $$z = i({\pi \over 2} + 2\pi n), \quad n \in \Bbb{Z}$$

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\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.

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$b$ should be $(2n+\frac{1}{2})\pi$. – 23rd Jan 26 '13 at 4:53
Good catch, @richard, thanks. In fact that is a typo of mine, which was meant to be $\,(4n+1)\pi/2)$ – DonAntonio Jan 26 '13 at 8:57
You are welcome! – 23rd Jan 26 '13 at 9:04