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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,

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    $\begingroup$ The upper half-plane maps to a certain strip. (The entire plane maps onto a larger strip). That real part is not positive. I didn't say you'd be done. Once you get straight what strip the logarithm maps the upper half-plane onto, figure out what part of the upper half-plane maps onto half that strip... $\endgroup$ – David C. Ullrich Jun 30 '15 at 0:41
  • $\begingroup$ Hi @DavidC.Ullrich, so I think Log(z) maps the UHP to the infinite strip, 0<y<$\pi$, and Quadrant I maps to the bottom half, 0<y<$\pi/2$, while Quadrant II of the z-plane maps to the upper half of this strip. Now I have to somehow make the strip semi-infinite and then cover $-\pi < y < \pi$ $\endgroup$ – User001 Jun 30 '15 at 1:38
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    $\begingroup$ True, but not the "half strip" I was asking about - what you say here doesn't help. Say $S$ is the strip $0<y<\pi$. Say $U$ is the semi-strip defined by $0<y<\pi, x>0$. So Log maps the upper half-plane onto $S$. Now the question is this: Log maps exactly what part of the upper half-plane onto $U$? $\endgroup$ – David C. Ullrich Jun 30 '15 at 17:22
  • $\begingroup$ Hmm... to have positive real part in the image, ln|z| must be positive, so |z| must be greater than 1. This means Log maps everything in the UHP that's outside of the upper half of the unit disk. This region maps to U, under the action of Log(z). What next? How can I handle the space inside of the upper half of the unit disk in the z-plane? Also still need my semi strip to cover -pi < y < pi. Thanks, $\endgroup$ – User001 Jun 30 '15 at 20:30
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    $\begingroup$ We're getting there. Absolutely right, Log maps the part of the upper half plane outside the unit disk to the semi-infinite strip you want. Let's say $S$ is the set of $z$ in the upper half-plane with $|z|>1$. Now the tricky part. Say $Q$ is the first quadrant and $\Pi^+$ is the upper half--plane. There exists a $\phi$ mapping $\Pi^+$ to itself that maps $Q$ to $S$. Hint: Saying $\phi$ maps $\Pi^+$ to itself says exactly that $\phi(z)=(az+b)/(cz+d)$ where $a,b,c,d$ are real and $ad-bc>0$. Hint: $S$ has "corners" at $1$ and $-1$, while $Q$ has "corners" at $0$ and $\infty$. $\endgroup$ – David C. Ullrich Jun 30 '15 at 20:41
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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.

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  • $\begingroup$ Hi Professor Israel, thanks so much for the clarification. I currently have an infinite strip, $0< v < \pi$, in the w-plane. Can I please get another hint on how to proceed? I need a semi-infinite strip, and then have it cover $-\pi < v < \pi$ $\endgroup$ – User001 Jun 30 '15 at 1:52
  • $\begingroup$ Hi Professor Israel, I'd like to think that I was very close :( my answer is f(z) = $2 Log(\frac {-\sqrt(z) - 1}{\sqrt(z) -1}) $, but I think it may be f(z) = $2 Log(\frac {-\sqrt(z) - i}{\sqrt(z) -i}) $ instead. Thanks so much for your help. I will start a new question on some basic mappings that are confusing me. $\endgroup$ – User001 Jul 1 '15 at 1:44
  • $\begingroup$ Hi @RobertIsrael, I just tried this problem again and am getting a slightly different answer. When I get myself to the region in the upper half plane, outside of the unit disk, I had originally rotated this space 180 degrees, and applied the Log(z) map, and multiplied by 2. But I think the rotation should only be 90 degrees - so that the target space is on the right half plane. What do you think? If this is correct, then I have my answer as 2Log($\frac{i\sqrt{z}-1}{\sqrt{z}-i})$ $\endgroup$ – User001 Jul 1 '15 at 6:10
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    $\begingroup$ If $z$ is in the upper half plane, $\sqrt{z}$ is in the first quadrant, and $ \dfrac{i\sqrt{z}-1}{\sqrt{z}-i}$ is in the left half plane outside the unit circle. You want to take a logarithm of that, but not the principal branch logarithm because you don't want a branch cut running through this region. $\endgroup$ – Robert Israel Jul 1 '15 at 6:29
  • $\begingroup$ Oh, hmm...I realized I rotated counterclockwise (positive direction). What if I rotated the region in the upper half plane, outside the unit circle, 90 degrees clockwise with the scalar -i? Then my answer becomes $2Log(\frac{-i\sqrt{z} + 1}{\sqrt{z} - i})$, keeping the principal branch of log. What do you think? There should be no issues with the branch cut along the negative real axis now, @RobertIsrael. $\endgroup$ – User001 Jul 1 '15 at 6:52

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