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What is the number of roots of $$z^2= \sin(z)$$ in $\left\{z\in\mathbb C\setminus\mathbb R\left||z|<2\right.\right\}$?

This task is related to the Argument principle and Rouché's theorem. Maybe I should estimate $f(z)=z^2$ and $g(z)= \sin(z)$ using an inequality that would hold when $|z|<2$.

I would be grateful if you provide an explanation, so I could learn how to solve similar problems.

gt6989b
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whatanhonor
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1 Answers1

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Consider $g(z)=\sin z-z^2$ and $D=\{z\in\mathbb{C}:\lvert z\rvert<2\}$. I claim that $$\bbox[10px,#ffd]{ \lvert \sin z\rvert<\lvert z\rvert^2 } $$ on $\partial D$. From this approach, if we let $z=x+iy$, then $$ \max_{\lvert z\rvert=2}\, \lvert\sin z\rvert=\max_{x^2+y^2=4}\,\left\lvert\sqrt{\cosh^2y-\cos^2x}\right\rvert $$ The last function has (on the constrained domain) two maxima at $(0,2)$ and $(0,-2)$, which is equivalent to $\pm2i$. Evaluating at these values yields $$ \lvert\sin(\pm2i)\rvert=\sqrt{\cosh(2)^2-1}\approx3.62 $$ So on $\partial D$, $$ \lvert\sin z\rvert\leq 3.63<4=\lvert z\rvert^2 $$ and our claim is justified. Now, by Rouche's theorem, $z^2$ and $g(z)$ have the same number of zeros (counted with multiplicity) in $D$ - namely 2.

Now, we consider the equation $$ \sin z=z^2 $$ for $z\in\mathbb R$ only. Since $|\sin(z)|\le 1$ for $z\in\mathbb R$, we must have $z\in[-1,1]$. Now we note that $z=0$ is a real solution. Since $$(0.1)^2<\sin(0.1) \text{ and } 1^2=1>\sin(1),$$ we know from the Intermediate value Theorem that there is another solution for some $0.1<z<1$.

Hence, both solutions in $D$ are real and it follows that there are no non-real solutions to $\sin(z)=z^2$ on $D$.

Maximilian Janisch
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Ken
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  • thank you very much! – whatanhonor Dec 24 '19 at 21:27
  • @whatanhonor No problem mate, this was a fun one! – Ken Dec 24 '19 at 21:29
  • @Kenny It is safer to say $|\sin(z)|\le 3.63$. Otherwise - great answer! – Maximilian Janisch Dec 25 '19 at 14:51
  • Also, I am not sure where you need the identity principle. If $f$ $g$ are two functions taking complex arguments and $f(x)=g(x)$ for some $x\in\mathbb R$ then surely also $f(z)=g(z)$ for $x=z\in\mathbb C$. No identity principle is required and in particular $f$ and $g$ need not be holomorphic for this to hold – Maximilian Janisch Dec 25 '19 at 14:55
  • @MaximilianJanisch I learned it in Gamelin as the uniqueness principle which was essentially a corollary to an analytic function having isolated zeroes. Given $f,g$ holomorphic on a domain $D$, if they are equivalent on a non-isolated set (here I took it as $\mathbb{R}$) then they are equal on the entire domain $D$ (here I took it as the disk of radius 2). Does that equivalence hold for non-holomorphic functions? If it does then I am pleasantly surprised. – Ken Dec 25 '19 at 16:18
  • @MaximilianJanisch Well actually a simple counter example. Consider the entire functions $e^z$ and $sinz$. They are certainly equal on the set ${k\pi/2:k\in\mathbb{Z}}$. However, this set is discrete. Therefore the two functions are not equal on the entire complex plane. Consider $z=0$ where they differ in value. – Ken Dec 25 '19 at 16:28
  • @MaximilianJanisch So we require a non-isolated set. And now consider $\lvert z \rvert$ and $\Re{z}$ which are equal on a non-isolated set (the real line). However, these functions are not holomorphic, and in particular are not equal off of the real line. – Ken Dec 25 '19 at 16:43
  • @Kenny You misunderstood my original point. Indeed, as you correctly state, the Identity Theorem of course does not hold for all functions $f,g:\mathbb C\to\mathbb C$ (but it does hold for all holomorphic functions). I will try to restate my original point (which is not about complex analysis but rather just about functions in general) in a more precise way: – Maximilian Janisch Dec 25 '19 at 17:42
  • $\newcommand{\v}{\vert_{\mathbb R}}$ Let $\Omega$ be any set and $f,g:\mathbb C\to\Omega$ any two functions. Let $$f\v:\mathbb R\to\Omega, x\mapsto f(x)$$ and $g\v$ be defined similarly. Then we have $$\bbox[15px,border:1px groove navy]{f\v(x)=g\v(x)\implies f(x)=g(x)}$$ for all $x\in\mathbb R$ – Maximilian Janisch Dec 25 '19 at 17:42
  • @Kenny In your case, we have $f(z)=\sin(z)$ and $g(z)=z^2$. What I am saying is that, because the functions are already defined on the entire $\mathbb C$ a priori, you don't need to use Identity Theorem in order to conclude that $\sin(z)-z^2$ has two real zeros. It is clear without Identity Theorem, because if $x\in\mathbb R$ and $\sin(x)-x^2=0$ for some $x\in\mathbb R$, then we have $\sin(z)-z^2=\sin(x)-x^2=0$ for $z=x\in\mathbb C$ (this is just a statement about functions, basically I am just saying that if $x\in\mathbb R$ and $z\in\mathbb C$, then $x^2=z^2$ and $\sin(x)=\sin(z)$) – Maximilian Janisch Dec 25 '19 at 17:47
  • @Kenny I hope this clears up the ambiguity. I am just noting that your argument doesn't require the identity principle. Of course, I would be happy to continue this discussion in chat if you disagree: https://chat.stackexchange.com/rooms/102565/non-real-roots-of-z2-sinz – Maximilian Janisch Dec 25 '19 at 17:50
  • @Kenny Sorry, in the second last comment I of course meant "if $x\in\mathbb R$ and $z\in\mathbb C$ with $x=z$, then $x^2=z^2$ and $\sin(x)=\sin(z)$" – Maximilian Janisch Dec 25 '19 at 17:59
  • @MaximilianJanisch Ah I see now, sorry about the misunderstanding. I completely agree with what you're saying. – Ken Dec 25 '19 at 18:04
  • @Kenny I have taken myself the right to edit your answer accordingly. Please feel free to rollback my edit in case you don't like it or to improve my edit and correct any mistakes I may have made. Also, merry christmas – Maximilian Janisch Dec 25 '19 at 18:36