# Thread: Find a complex number

1. ## Find a complex number

I'd like to check my result.
Find all the complex numbers $z$ such that $e^{i \overline z}= \overline {e^{iz}}$.
I've found that $z=x$, where $x \in \mathbb{R}$. So when $z$ is any real number.

2. Originally Posted by arbolis
I'd like to check my result.
Find all the complex numbers $z$ such that $e^{i \overline z}= \overline {e^{iz}}$.
I've found that $z=x$, where $x \in \mathbb{R}$. So when $z$ is any real number.

Also, consider $z=\pi/2$. Then $e^{i\overline{\pi/2}}=e^{i\pi/2}=i\neq -i = \overline{e^{i \pi/2}}$

3. EDIT. I made a mistake.

$e^{i\overline{z}} = e^{i(x-iy)}=e^{y+ix} = e^{y}e^{ix}= e^{y}\big(\cos(x)+i\sin(x)\big)$

$\overline{e^{iz}} = \overline{e^{i(x+iy)}} = \overline{e^{-y+ ix}} = \overline{e^{-y}e^{ix}} = e^{-y}\big( \cos(x) - i \sin(x)\big)$

so it seems to be for only $z= n \pi + 0i \ \ n= 0, \pm 1, \pm 2, ...$

4. Originally Posted by arbolis
I'd like to check my result.
Find all the complex numbers $z$ such that $e^{i \overline z}= \overline {e^{iz}}$.
I've found that $z=x$, where $x \in \mathbb{R}$. So when $z$ is any real number.
No. If z= x, a real number, $i\overline{z}= iz$ while $\overline{iz}= -iz$ In general, $e^{iz}\ne e^{-iz}$.

5. Originally Posted by Random Variable

$e^{i\overline{z}} = e^{i(x-iy)}=e^{y+ix} = e^{y}e^{ix}= e^{y}\big(\cos(x)+i\sin(x)\big)$

$\overline{e^{iz}} = \overline{e^{i(x+iy)}} = \overline{e^{-y+ ix}} = \overline{e^{-y}e^{ix}} = e^{-y}\big( \cos(x) - i \sin(x)\big)$

so it seems to be for only $z= n \pi + 0i \ \ n= 0, \pm 1, \pm 2, ...$
Okay, if $z= n\pi$, then $e^{i\overline{n\pi}}= e^{n\pi i}= cos(n\pi)+ i sin(n\pi)= (-1)^n$ while $e^{\overline{i n\pi}}= e^{-i n\pi}= cos(-n\pi)+ i sin(n\pi)= (-1)^n$ since cosine is an even function.

6. Originally Posted by HallsofIvy
Okay, if $z= n\pi$, then $e^{i\overline{n\pi}}= e^{n\pi i}= cos(n\pi)+ i sin(n\pi)= (-1)^n$ while $e^{\overline{i n\pi}}= e^{-i n\pi}= cos(-n\pi)+ i sin(n\pi)= (-1)^n$ since cosine is an even function.
So are you saying I'm right?

7. Originally Posted by HallsofIvy
Okay, if $z= n\pi$, then $e^{i\overline{n\pi}}= e^{n\pi i}= cos(n\pi)+ i sin(n\pi)= (-1)^n$ while $e^{\overline{i n\pi}}= e^{-i n\pi}= cos(-n\pi)+ i sin(n\pi)= (-1)^n$ since cosine is an even function.
Actually it is if $z= n\pi$, then $e^{i\overline{n\pi}}= e^{n\pi i}= cos(n\pi)+ i sin(n\pi)= (-1)^n$ while $\overline{e^{i n\pi}}= \overline{cos(n\pi)+ i sin(n\pi)} = cos(n\pi) - i sin(n\pi) = (-1)^n$

because in the right hand it is $\overline{e^{iz}}$ and not $e^{\overline{iz}}$

even if it is the same since

$\overline{e^{iz}} = \overline{e^{i(x+iy)}} = \overline{e^{-y}e^{ix}} = e^{-y}\\cos x - i \sin x)" alt="\overline{e^{iz}} = \overline{e^{i(x+iy)}} = \overline{e^{-y}e^{ix}} = e^{-y}\\cos x - i \sin x)" /> while

$e^{\overline{iz}} = e^{-i \overline{z}} = e^{-i (x-iy)} = e^{-y}\\cos x - i \sin x)" alt="e^{\overline{iz}} = e^{-i \overline{z}} = e^{-i (x-iy)} = e^{-y}\\cos x - i \sin x)" />