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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.