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Let $U=D_{1/2}(0)$. Show that there is a holomorphic function $h:U\to\mathbb C$ such that $$e^{h(z)}=1+z^5+z^{10}$$

My proof: Such a function exists if $f(z):=1+z^5+z^{10}$ has a logarithm on $U$, because in that case we can set $h:=\log f$. Now, $f$ has a logarithm iff $\frac{f'}{f}$ has an antiderivative on $U$.

Now I take $f_1(z):=z^{10}+z^5$ and $f_2(z):=1$. On $\partial U$ we have $|f_1(z)|<|f_2(z)|$ so by Rouché's theorem $f=f_1+f_2$ has no roots in $U$ and thus $\frac{f}{f'}$ is holomorphic which means $\oint_\gamma\frac{f}{f'}dz=0$ for every closed contour $\gamma$ in $U$. Thus the function has an antderivattive which finishes the proof.

Is this correct?

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Yes, it is correct, but you only need to prove that $f$ has no zeros on $D_{\frac12}(0)$ and then use the fact that that disk is simply connected. Every holomorphic function without zeros whose domain is simply connected has a holomorphic antiderivative.

Besides, you should have written $\frac{f'}f$ instead of $\frac f{f'}$.

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If $1+z^5+z^{10}=0$, then $z^5= -\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$, thus $|z|=1$.

Conclusion: the function $f(z)=1+z^5+z^{10}$ has no zeros in $D_{\frac12}(0)$.

$D_{\frac12}(0)$ is simply connected and the result follows.

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