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We have for example the problem: Show that all the roots of $f(z) = z^5 + 3z + 1$ are in the disk $|z| < 2.$

The solution to this is not so hard as we can let $g(z)=-z^5$, this has $5$ roots in the disk $|z|<2,$ then for some $z\in \Gamma$, $$|f(z)+g(z)|\le 3|z|+1=7<|g(z)|=32$$

Clearly by Rouché's Theorem, $f$ has $5$ roots inside $|z|=2.$

Then come the problem I am stuck in, unlike above which I solved.

Problem: Prove that $e^z+z^3$ has no root in $\{z:|z|<3/4\}$ and has three roots in $\{z:|z|<2\}$.

My attempt: Following the above method that I used, if we let some $f(z)=-z^3,$ then we have $g(z)=e^z+z^3$ which gives us in $|z|=2:$

$$|g(z)+f(z)|=|e^z|= e^x\le e^{2}<|f(z)|=2^3$$

Then by Rouché's theorem, $f$ and $g$ has the same number of roots, and since $f(z)$ has $3$ roots (counting multiplicity) therefore $g(z)$ must have $3$ roots inside the disk $|z|=2.$

Where I am stuck: I am stuck in trying to show the case for when there are no roots inside the disk $|z|=3/4.$ I am assuming we work with $e^z$ since $e^z$ has no roots anywhere as its never $0.$

I would appreciate the help.

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    $\begingroup$ For $|z|<3/4$ you have $|e^{z}|=e^{Re(z)}\in[e^{-3/4},e^{3/4}]$ and $|z^3|<(3/4)^3$. Compare $(3/4)^3$ and $e^{-3/4}$. $\endgroup$ – user553213 Apr 23 '18 at 11:34
  • $\begingroup$ Thank you for the reply, my question from this now is why is the absolute value of $e^z$ equivalent to $e^{\Re (z)}?$ $\endgroup$ – Aurora Borealis Apr 23 '18 at 11:52
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    $\begingroup$ $|e^z|=|e^{Re(z)}||e^{iIm(z)}|$, but $|e^{iIm(z)}|=\sqrt{(\cos(Im(z)))^2+(\sin(Im(z)))^2}=1$ by Pythagoras. $\endgroup$ – user553213 Apr 23 '18 at 11:59
  • $\begingroup$ So I thought about it and if we consider the case when $|z|=3/4$, then let $ (z)=-e^z$ then we have $|g(z)+f(z)|=|z^3|=(3/4)^3<|f(z)|=e^{3/4},$ and therefore $f(z)$ has same number of roots as $g(z)$ and since $f(z)$ has no roots in $\mathbb{C}, g(z)$ also has no roots in the disk $|z|=3/4.$ Is this correct? $\endgroup$ – Aurora Borealis Apr 24 '18 at 3:26

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