I want to map the upper half plane, y>0, conformally onto the semi-infinite strip u>0, $-\pi < v < \pi$ in the w-plane.
I then studied the complex logarithm, and noticed that the principal branch, Log(z), maps every point z in $C – R^- \bigcup {0}$ to w = ln|z| + iArg(z), where Arg(z) ranges from $-\pi$ to $\pi$.
So, the images of the whole plane, minus the negative real axis (and 0), have positive real part and imaginary part between $-\pi$ to $\pi$.
Is my work done? Can I conclude that the mapping from the UHP to the semi-infinite strip is just $$w = Log(z)?$$
Something seems a little off, since all of C (minus the negative real axis) maps to the horizontal strip.
Thanks,
Best Answer
A point in the UHP is $z=r e^{i\theta}$ with $r > 0$, $0 < \theta < \Pi$. $w = \text{Log}(z)$ maps this to $w = \log(r) + i \theta = u + i v$ with $u = \log(r) \in (-\infty, \infty)$, $v = \theta \in (0,\pi)$. Note that the logarithm of a positive number could be negative. So you have the infinite strip, not a semi-infinite strip.
EDIT:
Hint: half-plane $\implies$ quarter-plane $\implies$ half-disk $\implies$ semi-infinite strip.