# Homework Help: Determine f(S)

1. Jan 19, 2014

1. The problem statement, all variables and given/known data
Determine $f(S)$ where $f(z)=e^{\frac{1}{z}}$ and $S=\{z:0<|z|<r\}$.

*Edit: The function f is defined as $f:\mathbb{C}\rightarrow\mathbb{C}$.

3. The attempt at a solution
I am a little confused as to what this problem is asking me to do. What I did was:

Let $z=|z|e^{i\theta}$. Then
\begin{align*} f(z)=e^{\frac{1}{z}}=e^{\frac{1}{|z|e^{i\theta}}}=e^{\frac{\cos\theta-i\sin\theta}{|z|}} =e^{\frac{\cos\theta}{|z|}}e^{-\frac{i\sin\theta}{|z|}} =e^{\frac{\cos\theta}{|z|}}\left(\cos(\frac{\sin\theta}{|z|})-i\sin(\frac{\sin\theta}{|z|})\right). \end{align*}

Can I then say that $f(S)=e^{\frac{\cos\theta}{|z|}}\left(\cos(\frac{\sin\theta}{|z|})-i\sin(\frac{\sin\theta}{|z|})\right)$ for $S=\{z:0<|z|<r\}$ or is there something more to this problem that I am not seeing?

Last edited: Jan 19, 2014
2. Jan 19, 2014

### Dick

Yes, you are kind of spinning around. $e^{\frac{1}{z}}=e^{\frac{1}{|z|}}e^{-i\theta}$. If 0<|z|<r where is 1/|z|?

3. Jan 19, 2014

If $0<|z|<r$ then we have $\frac{1}{r}<\frac{1}{|z|}$ and $\frac{1}{|z|}\rightarrow\infty$ as $|z|\rightarrow 0$ but $0<|z|$ so we can safely say $\frac{1}{|z|}<\infty$.

4. Jan 19, 2014

### Dick

Ok, so 1/r<1/|z|<∞. S is a disk of radius r with a hole at z=0. What's f(S)?

Last edited: Jan 19, 2014
5. Jan 19, 2014

If $f(z)=\frac{1}{z}$, $f(S)$ would be a disk with a hole inside it centered at the origin with radius $\frac{1}{r}$ but in this case, the function is the exponential.

I think $f(z)=e^z$ maps z to a circle on the complex plane of radius Re(z) so I'm tempted to say f(S) is a mess of circles each with radius larger than $Re(\frac{1}{r})$.

Edit** When I say a mess of circles, in my mind I picture a circle of radius $\frac{1}{r}$ that is expanding outwards.

6. Jan 19, 2014

### Dick

f(z) isn't 1/z, if it were f(S) would be everything outside of a disk of radius 1/r. it's e^(1/z). Can you rethink that a little?

Last edited: Jan 19, 2014
7. Jan 19, 2014

I can't seem to picture it. In class I was shown that if z=x and Im(z)=0 then e^z was a circle and if Re(z)=0 and Im(z)=y then e^y was a vector that pointed outwards from the origin at an angle of y. Combining these together all I can see is two circles, one inside the other bounding the area between them.

8. Jan 20, 2014

### Dick

Oh, heck. I'm completely on the wrong track here and I apologize. What I said before is completely wrong. Utterly and completely wrong. Please ignore everything I've said. I have to give this a serious rethink. Sorry again.

Last edited: Jan 20, 2014
9. Jan 20, 2014

I'm confused. I don't see the point in this exercise if the work I was doing before your input was correct.

10. Jan 20, 2014

### Dick

I think they probably want you use Picard's theorem about essential singularities.

11. Jan 20, 2014

This problem comes from the first chapter in the textbook which is an introduction complex analysis. Picard's theorem comes in chapter 4.

Do you know if there is any way to parametrize S without $x=|z|\cos\theta$, $y=|z|\sin\theta$?

Last edited: Jan 20, 2014
12. Jan 20, 2014

### Dick

Don't parameterize at all. It's not going to be helpful, it's nasty any way you do it. Take a more direct approach. 1/z maps the disk to the exterior of the disk. Use what you know about the exponential map to say what f(S) must be. It's going to be all of the values you can get by applying exp to the exterior of the disk. The conclusion is actually pretty easy, if you think about it.

Last edited: Jan 21, 2014