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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!

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1 Answer 1

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

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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

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