Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I just need some help with the penultimate question of my coursework:

Let $w=f(z)=\coth(z/2)$. Show that $w=f(z)=h(C) = (C+1)/(C-1)$ where $C=g(z)=e^z$. Find the image of the given points, boundary and region under $C=g(z)=e^z$, in the $\mathbb{C}$-plane (complex plane).

Here is the curve.

Hence find the image of these points, boundary and region under $w=f(z)=\coth(z/2)$

Thanks guys, any help will be appreciated!

share|cite|improve this question

Hint: $\coth(z/2) = \frac{\cosh(z/2)}{\sinh(z/2)}$. Write these in terms of exponentials, and multiply numerator and denominator by $\exp(z/2)$.

share|cite|improve this answer
Aha! Thanks, this gives the correct for C+1/C-1.... Now I'm stuck, do I put the values of ipi and 0 into f(z)? Doing this gives 0/1 and 1/0 so I don't see how this gives an image – bany Mar 22 '12 at 17:50
Right so... I've shown the second part of the question, I managed to end up with the C plane showing a semi-circle between -1 and 1... Now for the last part of the question, f(0)= infinity and f(ipi) = 0, so what does this even make an image to? – bany Mar 22 '12 at 18:31
If $z$ is in the strip $0 \le \text{Im} z \le \pi$, $C = e^z$ is in the upper half plane $\text{Im}(C) \ge 0$ with $0$ removed. Now $C \to \frac{C+1}{C-1}$ is a Möbius transformation. What does it do to points on the real axis? What does it do to the upper half-plane? – Robert Israel Mar 22 '12 at 20:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.