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I have some trouble with the following question:

Let $$f(z)=\frac{\sin z}{z},\quad\text{for }z\in\mathbb{C}.$$ What is the maximum of $f$ in the closed unit disc $$D:=\{z\in\mathbb{C}:|z|\leq 1\}?$$

Of course, we can apply the Maximum Modulus Principle to get that $$\max_{z\in D}|f(z)|=\max_{|z|=1}\frac{|\sin z|}{|z|}=\max_{|z|=1}|\sin z|$$ But then I don't see any easy way to find the maximum of $\sin z$ on the unit circle. I ploted the graph and I found that it is $$|\sin i|=\frac{e-e^{-1}}{2}$$ Here is my attempt to prove it:

Let $F(\theta)=|f(e^{i\theta})|^2$. Then, \begin{align} F(\theta) &= |\sin(\cos\theta+i\sin\theta)|^2 \\ &= |\sin(\cos\theta)\cosh(\sin\theta)+i\cos(\cos\theta)\sinh(\sin\theta)|^2 \\ &= \sin^2(\cos\theta)\cosh^2(\sin\theta)+\cos^2(\cos\theta)\sinh^2(\sin\theta) \\ &=\sin^2(\cos\theta)\cosh^2(\sin\theta)+(1-\sin^2(\cos\theta))\sinh^2(\sin\theta) \\ &=\sin^2(\cos\theta)\big(\cosh^2(\sin\theta)-\sinh^2(\sin\theta)\big)+\sinh^2(\sin\theta) \\ &= \sin^2(\cos\theta)+\sinh^2(\sin\theta) \end{align} Hence, \begin{align} F'(\theta) &= -2\sin(\cos\theta)\cos(\cos\theta)\sin\theta+2\sinh(\sin\theta)\cosh(\sin\theta)\cos\theta \\ &=\sinh(2\sin\theta)\cos\theta-\sin(2\cos\theta)\sin\theta \end{align} which vanishes whenever $$\theta=\frac{n\pi}{2},\quad n\in\mathbb{Z}.$$ Then, we try and find $$F(0)=\sin^2 1,\quad F(\pi/2)=\sinh^2 1,\quad F(\pi)=\sin^21,\quad F(3\pi/2)=\sinh^21$$ But $\sin^21<\sinh^21$ (which I am not sure how to justify with only pen and paper) so the maximum is with $\theta=\pi/2$ or $3\pi/2$. This gives the desired result.

This seems overly complicated for this kind of problem. Is there a better way?

Spenser
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3 Answers3

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The Taylor series for $\frac{\sin(z)}{z}$ has all its terms real and positive when $z=i$, and the corresponding terms are the same in magnitude for any $z$ on the unit circle, so the maximum has to occur for $z=i$.

universalset
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    This answer is great because it enlightened me to a new way of looking at the behavior holomorphic functions. I am sure this kind approach will be useful for me later. Many thanks ! – Spenser Dec 05 '13 at 21:34
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Use the series expansion (if we can use): $$e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!},\qquad\sin z=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)!}z^{2n-1},\forall z\in \mathbb{C}.$$ So, when $|z|=1$, $$|\sin z|=\left|\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)!}z^{2n-1}\right| \leq\sum_{n=1}^{\infty}\frac{1}{(2n-1)!}.$$ Obviously, we have $$|\sin(\pm i)|=\frac{e-e^{-1}}{2}=\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{1}{n!}-\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\right)=\sum_{n=1}^{\infty}\frac{1}{(2n-1)!}.$$ So $$\max_{|z|=1}|\sin z|=|\sin (i)|=\frac{e-e^{-1}}{2}.$$

Riemann
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Here is an approach, since $z=x+iy$, then

$$ \max_{|z|=1}|\sin(z)|= \max \left\{\sqrt{\cosh(y)^2-\cos(x)^2}\right\},\quad x^2+y^2 = 1. $$

Now, we need to maximize the above function over the unit circle $x^2+y^2= 1$. We can use optimization techniques (see multivariable calculus). Here is an answer by Maple

$$ \rm {Max:} \sim 1.1752, (x=0,y=1)$$

Note: To make the calculations easier, you can just find the maximum of the function

$$ \cosh(y)^2-\cos(x)^2, \quad x^2+y^2 = 1, $$

then take the square root of the answer.

Mhenni Benghorbal
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