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Are there any zeros of multiplicity greater than or equal to 2 in the unit disk for the function $f(z)=4z^{10}-e^{z}$?

So, first I set $g(z)=4z^{10},$ then observed that on $\partial D$ we have $|f-g|=|e|\le|g|=|4|.$ Rouche's theorem then implies $f$ and $g$ have same number of zeros inside $D,$ which is 10. Then how should I prove the multiplicity of zero?

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The complex number $z_0$ is a zero of $f$ of multiplicity greater than or equal to 2 iff $f(z_0)=0$ and $f'(z_0)=0$. Now $$\begin{cases} 4z^{10}-e^{z}=0 \\ 40z^{9}-e^{z}=0 \end{cases}\implies 4z^9(z-10)=0\implies z=0, z=10$$ and neither of them are acceptable solutions.

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  • $\begingroup$ Thanks, that makes sense! $\endgroup$ – user456649 Jul 13 '17 at 16:26
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A multiple zero of $f(z)$ would also be a zero of $f'(z)$, and thus a zero of $f(z)-f'(z)=4 z^9(z-10)$, but neither $0$ nor $10$ is a zero of $f(z)$, so all zeroes are simple.

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  • $\begingroup$ Thanks, it helped me to work it out! $\endgroup$ – user456649 Jul 13 '17 at 16:27

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