The image of a horizontal line Im(z)=pi/4 for the complex number f(z)=e^z

Question

I tried to plot the image of the line $Im(z)=\pi/4$, for the complex number $f(z)=e^z$.

I wrote first the complex number as w=u+iv:

$w=e^x\cdot e^{-iy}$

$Im(z)=\pi/4$ means that y in the equation is $\pi/4$, so that gives:

$Im(z)=e^{-i\pi/4}=\cos(\pi/4)-i\sin(\pi/4)$

then

$w=e^x(\cos(\pi/4)-i\sin(\pi/4))=e^x\cos(\pi/4)-e^xi\sin(\pi/4)$

Plotting this in a w-plane , I thought it would be the line at $-i\sin(\pi/4)=-\sqrt{2}/2$, but I actually just replotted the z-geometry again. See image below.

How do I do this right?

Thanks

Answer 1

If $z=x+\frac\pi4i$, then$$\exp(z)=e^x\left(\frac{1+i}{\sqrt2}\right).$$Since $\{e^x\mid x\in\Bbb R\}=(0,\infty)$, $\left\{\exp\left(x+\frac\pi 4i\right)\,\middle|\,x\in\Bbb R\right\}$ is the open ray with origin in $0$ and passing through $\frac{1+i}{\sqrt2}$.