I'm reposting with my attempts added as my previous question was closed.
Here is a problem from a complex analysis exam. These to seem to be related, but I can't see how to solve them.
a) Let $c\in \mathbb{R}$ be such that $c>e$. Show that there exists exactly one $z \in \mathbb{D}$ such that
$$e^z=cz$$ and that $z$ is a real number.
b) Let $z \in \mathbb{C}$ be a complex number satisfying $$e^z=ez$$ Show that $Rez\ge 1.$
The two tasks are separate [ that is $z$ in a) has nothong to do with $z$ in b) ] and $\mathbb{D}$ denotes unit disc.
Attempt:
For b) I've tried substituting $z=x+iy$ which after comparing real and imaginary parts yields:
$$e^x\cos{y}=xe ; e^x\sin{y}=ye$$
and assuming $x<1$ I tried to find contradiction but I eventually couldn't find it.
Both a) b) also look to me a bit like Schwarz Lemma thing because they have the form of $f(z)=\lambda z$ but I didn't manage to find proper function $f$ so that the Lemma's assumptions are satisfied.
Could you give me some hints or solutions?
Best Answer
For 1) apply Rouche's Theorem. Let $f(z)=e^{z}/c$. Then, for $|z|=1$, we have $|(f(z)-z)+z|=|e^{z}/c|< 1=|z|$ so $f (z)-z$ has the same number of zeros in the unit disk as $z$,namely $1$.
For the second part let $z=x+iy$ ($x$ and $y$ real). Then $e^{x}\sin y =ey$. Since $|\sin y |\leq |y|$ we get $e^{x} \geq e$ and hence $x \geq 1$.