What is the image of a unit circle under the map $f(z)=1+z^2$?
I am stuck on the following problem:
The image of a unit circle under the map $f(z)=1+z^2$ is :
again the same unit circle
another circle with different centre but same radius
another circle with same centre but different radius
not a circle
None of the above
My Try: Consider the unit circle $|z|=1$. If we write this as $ \left\{ e^{i\theta} : 0 \leq \theta \leq 2 \pi \right\} $, then, its image under $z \mapsto {1+z^2} $ is readily seen to be $$ \{ 1+e^{2i \theta}=2 (\cos \theta )e^{i \theta} : \theta \in [0,2\pi]\} $$
(Using $e^{i\theta} + e^{-i\theta} = 2\cos(\theta)$)
Now, I am not sure how to progress further. Can anyone explain?
$\endgroup$ 52 Answers
$\begingroup$$z^2$ maps the unit circle onto itself. It actually maps the upper unit semicircle onto the whole unit circle and does the same with the lower unit semicircle. Then +1 shifts it to the right. So you get the circle $|z-1| = 1$.
$\endgroup$ $\begingroup$Adding to what Paul said, we have that $(e^{i\theta})^2=e^{i2\theta}$, with modulus 1, i.e., a point in the unit circle centered at the origin. Then we translate by $1$, to get the image set {$1+e^{i2\theta} $}. And every point is hit, i.e., the map is a bijection of $S^1$ to itself; $e^{i\theta_0}=e^{2(i\theta_0/2)}$. You can see the map is also an injection.
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