# Image of a complex function

1. Nov 5, 2012

### LikeMath

Let $f(z)=z+\frac{1}{z}$, the question is to find the image of this function on $|z|>1$.

To do so, I tried to find the image of the unit circle which is the interval [-2,2] and so I could not determine our image.

If also we tried to find the image of f we get
$f(re^{i\theta})=u+iv$
where
$u(re^{i\theta})=(r+\frac{1}{r})\cos \theta$
and
$v(re^{i\theta})=(r-\frac{1}{r})\sin \theta$
with r>1.

2. Nov 5, 2012

### haruspex

No, $f(re^{iθ})=re^{iθ}+e^{i(-θ)}/r$
At r=1 this comes to the same, so the image of |z|=1 is indeed the [-2,2] interval of ℝ. But for |z| > 1 there will be an imaginary component.
Looks to me like the image is still the whole of ℂ. As you increase r, the image becomes a series of concentric ellipses. A proof might be derivable by showing there are always real solutions to x+iy = z + 1/z.