# Show existence of holomorphic function $h$ such that $e^{h(z)}=1+z^5+z^{10}$

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?

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 $$\frac f{f'}$$.
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.