Here is my question – it is an example sheet question, completely non-examinable:
Show that the equation $ z \sin(z) = 1 $ has only real solutions.
[Hint: Find the number of real roots in the interval $[-(n+1/2)\pi, (n+1/2)\pi$ and compare with the number of zeros of $ z \sin(z) – 1 $ in a square box $\{|\Bbb Re(z)|,|\Bbb Im(z)| < (n+1/2)\pi\}$.]
I'd be most grateful if someone were able to give me a suggestion as to how to do this! (I realise that there is a hint, but I can't work out the real roots either!
If I divide through by $z$, then I get $ \sin(z) = 1/z $, but we known that $\sin(z)$ has a Taylor series, ie no principal part of the Laurent series. Does this not mean that I am trying to equate
$$ \sum_{r=0}^\infty {(-1)^r z^{2r+1} \over (2r+1)!} = {1 \over z} \,?$$
Thanks! 🙂
Best Answer
Hint. It can be handled with Rouche's Theorem.
Let $f(z)=z\sin z$ and $g(z)=z\sin z-1$. It suffices to show that $$ \lvert f(z)\rvert >1, $$ for every $z=x+iy$, such that $|x|+|y|=\pi(n+\tfrac{1}{2})$.
Once you show this we are going to have that $$ \lvert f(z)-g(z)\rvert<\lvert f(z)\rvert $$ and hence $f$ and $g$ are going to have the same number of roots in: $$ R=\big\{x+iy: \lvert x\rvert+\lvert y\rvert<\pi(n+\tfrac{1}{2})\big\}. $$