This is a problem from a past qualifying exam in complex analysis. I'm working through these to study for my own upcoming qual. For this question, I think my proof is fairly straightforward, but I'd like to know whether or not it is correct and complete. I'm also interested in other ways of answer the question. Thanks!


Find how many solutions (counting multiplicity) the equation $\sin z = ez^4$ has on the unit disk $|z|<1$. Justify your answer.

My Solution:

Let $g(z) = \sin(z)$ and $f(z) = -ez^4$ and consider these functions on the unit circle $|z|=1$. We will show by Rouche's Theorem that since $|g(z)|<|f(z)|$, then $f(z)+g(z)$ has the same number of zeros inside the unit circle as $f(z)$ counting multiplicities, thus there are four solutions to the given equation.

First, we need to show that $|\sin(z)|<|e|$. We have $$ \begin{align*} |\sin(z)| &= \left\vert \frac{e^{iz} - e^{-iz}}{2i}\right\vert \\ &= \frac{1}{2} |e^{i(x+iy)} - e^{-i(x+iy)}| \\ &\leq \frac{1}{2}(|e^{i(x+iy)}| + |- e^{-i(x+iy)}|)\\ &= \frac{1}{2}(e^{-y} + e^y) \end{align*} $$

On $|z|=1$, we have $|y|\leq 1$, thus $$ |\sin(z)| \leq \frac{1}{2}(e^{-y} + e^y) \leq \frac{1}{2}(e^{1}+e^{-1})< \frac{1}{2}(2e) = e. $$

Now, we have that when $|z|=1$ $$ |\sin(z)|<e = e|z|^4 = |-ez^4|, $$

thus $|g(z)|<|f(z)|$, and by Rouche's Theorem, $f(z)+g(z)$ has the same number of zeros inside the unit circle as $f(z)$. Since $f(z) = -ez^4$ is a polynomial, we know by the fundamental theorem of algebra that it has exactly four roots counting multiplicity. Thus, $$ f(z) + g(z) = \sin(z) - ez^4 $$

has exactly four roots and $\sin(z) = ez^4$ has exactly four solutions. $\blacksquare$


2 Answers 2


There is just one tiny correction I have: if $|y|\leq 1,$ it doesn't follow that $e^{-y}\leq e^{-1};$ rather, its true that $e^{-y}\leq e.$ Note this does not affect your proof at all, which is otherwise seems correct!

  • $\begingroup$ Hmm. I'm concerned that it might actually affect the proof... I need strict inequality here such that $e^{-y} + e^y < 2e$ for $|y| \leq 1$. I know this is true, but I wonder if there's another straightforward way to show this? $\endgroup$
    – Serafina
    Jul 16, 2022 at 18:11
  • 1
    $\begingroup$ You know it's true for $|y|<1$ and for the cases when $y=\pm 1$ we have that $e+e^{-1}<2e$ so all should be good? $\endgroup$
    – PCeltide
    Jul 16, 2022 at 18:19
  • $\begingroup$ Ah, yes. True. I was forgetting that $y$ here is just a real number so $|y|$ just equals $\pm 1$. Thanks. $\endgroup$
    – Serafina
    Jul 16, 2022 at 21:47

You have $$|g(z)|=|\sin(z)| = |\frac{e^{iz}-e^{-iz}}{2i}| = |\frac{e^{2iz}-1}{2ie^{iz}}|\leq \frac{1}{2}\bigg(\frac{|e^{2iz}|+1}{|e^{iz}|}\bigg)\\\leq \frac{|e^{2i(\cos(\theta)+i\sin(\theta))}|+1}{2|e^{i\cos(\theta)-\sin(\theta)}|}\leq \frac{|e^{-2\sin(\theta)}\cdot e^{2i\cos(\theta)}|+1}{2\cdot e^{-\sin(\theta)}}\\\leq \frac{(e^{-2\sin(\theta)}+1)}{2\cdot e^{-\sin(\theta)}}\leq \frac{e^{2}+1}{2e}<e$$ where $z=\cos(\theta)+i\sin(\theta)\,0\leq\theta<2\pi$ is an arbitrary point on the unit disc.

After this, what you have done with Rouche's theorem is fine and you can conclude that the number of roots is $4$.


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