Contraction of g

Question

Let $$g:\mathbb{C}->\mathbb{C}, g(z) = e^{iz} $$ Find the contraction of g. My work:

$$|g'(z)|<1 \implies calculations \implies |e^{iz}|<1 $$ What is this (can it be a circle?). I just don't see it.

Answer 1

First of all a contraction is not just a function that satisfies $|\frac{d}{dz}g(z)|< 1$ but it has to, for a certain $k\in[0,1)$ satisfy $|\frac{d}{dz}g(z)|< k$. In fact a function whose derivative approaches $1$ without reaching it, like $x-log(x)$ isn't a contraction.

Now to answer your question we want to find all the $z\in \mathbb{C}$ s.t. $|ie^{iz}|<k$ for a certain $k\in [0,1)$. If we write $z=x+iy$ we get

$$|ie^{i(x+iy)}|<k \iff |e^{-y}|<k \iff e^{-y}<k \iff -y<ln(k)$$

So the all $z$ that satisfy $Im(z)>ln(k)$ will do.

Note that if $k=0$ then there are no solutions as $e^{x}>0 $ $\forall x$