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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! :)

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    $\begingroup$ The real roots occur when the two graphs in this picture cross: i.imgur.com/lYga94H.png . It's easy to see that there is one real root between adjacent extrema of the $\sin$ function (except two between $-\pi/2$ and $\pi/2$). You don't need to find the exact values. You only need to know how many there are in the given interval. Then try evaluating a contour integral around the square box to see what you can find out about the number of complex roots. If the question is correct, the number of complex roots will match the number of real roots you counted. $\endgroup$
    – Steve Kass
    Mar 12 '14 at 16:22
  • $\begingroup$ Thanks. :) You said to evaluate the contour integral around the square box - how does this determine the number of roots (from the Residue Theorem?)? $\endgroup$
    – Sam OT
    Mar 12 '14 at 17:52
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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: $$ R=\big\{x+iy: \lvert x\rvert+\lvert y\rvert<\pi(n+\tfrac{1}{2})\big\}. $$

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  • $\begingroup$ Thanks for that hint. I should have mentioned that it was in the Rouche's Theorem area of the sheet. What does $|f(z)| > 1$ say? It doesn't say that $|f(z)| > |g(z)|$, does it? Also, you haven't said anything about the number of real roots of $g$ in the given interval? (I assume that you show that the number is the same, implying only real roots.) $\endgroup$
    – Sam OT
    Mar 12 '14 at 17:48
  • $\begingroup$ And do you mean $ |x| = |y| = \pi (n+1/2) $, not $ |x| + |y| = \pi (n+1/2) $ (since the sum isn't a square box, but a circle? $\endgroup$
    – Sam OT
    Mar 12 '14 at 17:49
  • $\begingroup$ @SmileySam: See update. $\endgroup$ Mar 12 '14 at 19:27
  • $\begingroup$ @Yiogos Ok, so $ z \sin(z) $ and $ z \sin(z) - 1 $ have the same number of zeros in your defined $R$. I don't really know what your $R$ represents - certainly not the square box $\{|\Bbb Re(z)|,|\Bbb Im(z)| < (n+1/2)\pi\}$... $\endgroup$
    – Sam OT
    Mar 12 '14 at 19:34
  • $\begingroup$ Also I don't see how it relates the number of real and complex roots... sorry! $\endgroup$
    – Sam OT
    Mar 12 '14 at 19:35

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