This is an old qual problem. I consider the function defined by $f(z)=14z^{100}-5e^z$ and apply Rouche's Theorem. Let $g(z)=14z^{100}$. Then for $z$ on the boundary of the unit disc, $\vert f(z)-g(z)\vert=\vert 5e^z\vert<14=\vert g(z)\vert$. Since $g$ has 100 zeros in the unit disc, counting multiplicities, $f$ also has 100 solutions in the disc. However, just because $g$ has a zero of order 100 doesn't mean $f$ does (WolframAlpha gives distinct solutions http://www.wolframalpha.com/input/?i=14z%5E100+%3D+5e%5Ez). How do I determine the multiplicities of the zeros. Wouldn't I have to find the location of each root? I am not sure if I have the correct machinery to tackle this part.


1 Answer 1


If $a$ is a zero of $f$ then the multiplicity is how many derivatives of $f$ vanish also at $a$.

We have that $f'(z)=1400z^{99}-5e^z$.

Then $f(z)=f'(z)=0$ implies that $14z^{100}=1400z^{99}$. This can only happen if $z=0$ or $z=100$. From these the only inside the unit disc is $z=0$. But $z=0$ is not a solution of $f(z)=0$.

Since in the unit disc it never happens that $f$ and $f'$ vanish simultaneously then all the zeros there must be simple.


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