# Finding what a line with angle $\theta$ maps to under $f(z)=\frac{1}{2}(z+\frac{1}{z})$

I am trying to do an exercise from my tutorial in a complex analysis course, the T.A didn't have time to do it but he did mention what the answer is.

The problem is to find the image of a line with angle $\theta$ under $f$, which I understand as to find $$f(\{re^{i\theta}|\, r\in\mathbb{R}\})$$

where $$f(z)=\frac{1}{2}(z+\frac{1}{z})$$

I have calculated that $$f(re^{i\theta})=\frac{1}{2}\cos(\theta)(r+\frac{1}{r})+i(\frac{1}{2}\sin\theta(r-\frac{1}{r}))$$

From here I have tried to write $$A=\frac{1}{2}\cos(\theta),B=\frac{1}{2}\sin(\theta)$$ and to find some relation between $A(r+\frac{1}{r})$ and $B(r-\frac{1}{r})$ (and maybe using something like $A^{2}+B^{2}=\frac{1}{4}$ ).

I did not manage to find any relation that I can understand as something geometrical, the T.A have mentioned that the answer is a hyperbola.

Can someone please help me out and point me in the right direction ?

-

## 1 Answer

Hint: You already stated that $f( re^{i\theta}) = \left[ \frac {1}{2} \cos \theta (r+ \frac {1}{r} )\right] + i \left[ \frac {1}{2} \sin \theta ( r - \frac {1}{r} ) \right ] = X+iY$

What can you say about

$$\frac {X^2}{ \cos^2 \theta} - \frac {Y^2}{\sin ^2 \theta} ?$$

-
Yes, I was just writing that up. +1 –  coffeemath Feb 9 '13 at 1:14