How to show that $ze^z$ is univalent on the unit disk? How can I show that the analytic function $z \mapsto ze^{z}$ is univalent (i.e. injective) on the unit disk $\mathbb{D} = \{z \in \mathbb{C} : |z|<1\} ?$ I can show it on the real interval $]-1,1[$ using increasing properties but not in the complex setting.
Please, if possible, suggest me elementary methods (not involving Lambert functions).
 A: It is enough to show that $f$ is univalent on the boundary of the unit disc
($C$ is the unit disc, $f(C)$ is then an analytic Jordan curve $J$ so by the argument principle if $w$ is not in $J$ the number of zeroes of $f(z)-w$ in the unit disc is given by the winding number which is $0$ if $w$ is in the outer domain of $J$ and $\pm 1$ in the inner domain; so in the open unit disc $f$ assumes no value in the outer domain of $J$ and precisely one in the inner; since $f$ is an open map, $f(\mathbb D) \cap J$ is then empty so we are done, $f$ is $1-1$ and onto the interior of $J$)
So if $z=e^{it}, w=e^{i\theta}, f(z)=f(w)$, where $-\pi < \theta,t \le \pi$, we get by taking absolute values $\Re z=\Re w$ so $\cos t =\cos \theta$ hence $t = \pm \theta$. If $t=\theta, z=w$, while in the second case $w=\bar z$ so $\bar f(z)=f (\bar z)=f(w)=f(z)$ (noting that $ze^z$ has real Taylor coefficients at zero) implies $f(z)$ real.
Now $f(z)=e^{it}e^{\cos t+i \sin t}=e^{\cos t+i(\sin t +t)}$, so one gets $\sin (\sin  t+t)=0$ so $\sin t +t =k\pi$. Now $t=0, \pi$ imply again $z=w$ since then $-t=t$ modulo $2\pi$ (or easier $z$ real) so assume by taking $-t$ if needed that $0<t<\pi$ and then $0< \sin t +t <2\pi$ so we need to have $\sin t+t=\pi$ and that is impossible since by elementary calculus (non-negative derivative) the function $\sin t+t$ is increasing on $[0,\pi]$ and achieves $\pi$ precisely at the end $t=\pi$.
Hence $f(z)=f(w), |z|=|w|=1$ implies $z=w$ so $f$ is univalent on the unit circle, hence on the unit disc as shown above
