# Sketch the image under $w = \mathrm{Log} z$, with $\mathrm{Re}z > 0$

I am trying to sketch the image under $w = \mathrm{Log} z$, with $\mathrm{Re}z > 0$, but my result is different from the result in the solutions manual.

So, to my knowledge, $\mathrm{Log}z = \mathrm{log}|z| + i \mathrm{Arg}z$. So the points in the $w$-plane will have real part $\mathrm{log}|z|$ and imaginary part $\mathrm{Arg}z$. So the real part can be infinitely small or infinitely large or anything in between. Here, the solutions agree with me. But for the imaginary part, with the restriction that $\mathrm{Re}z > 0$, it should be that $-\pi/2 < \mathrm{Arg}z < \pi/2,$ shouldn't it? So the imaginary part of the points $w$ should be bounded between $-\pi/2$ and $\pi/2$? In the solutions manual, the sketch includes the whole $w$-plane, with no such restriction on the $y$-axis. This confuses me and I'm thinking that maybe I have misunderstood something. Any ideas?

Well, $w=\log z$ or $e^{u+iv}=x+iy$ shows that \begin{cases} e^u\cos v=x>0\implies -\dfrac{\pi}{2}<v<\dfrac{\pi}{2},\\ e^u\sin v=y. \end{cases} so the image is the horizontal strip $\{u+iv:\,-\dfrac{\pi}{2}<v<\dfrac{\pi}{2}\}$.