# Maximum of $|\sin(z)|$ as $\{z: |z| \leq 1 \}$

Maximum of $|\sin(z)|$ as $\{z: |z| \leq 1 \}$

So according to the Maximum Principle, the maximum is when $|z|=1$. I tried using the fact that $\sin z= \dfrac {e^{iz}-e^{-iz}}{2i}$, but didn’t know how to continue from there. Setting $z=e^{it}$ doesn’t help much.

I saw this question was answered before, however the answer uses hyperbolic functions ($\sinh, \cosh$) which I haven’t actually studied.

Thanks in advance for any assistance!

#### Solutions Collecting From Web of "Maximum of $|\sin(z)|$ as $\{z: |z| \leq 1 \}$"

Here’s an alternative solution, which uses a little bit of information about hyperbolic functions. It’s not as slick as KittyL’s solution, but I hope you learn something.

(I’ll hope you permit me to say something about hyperbolic functions, since (1) it’s good to know something about hyperbolic functions in complex analysis and (2) this solution does not use that many properties of hyperbolic functions. And anyway, for complex analysis purposes trigonometric and hyperbolic functions are basically the same.)

So here goes. We’d like an expression for $\sin z$ that makes it clear what the real and imaginary parts are, so that we can figure out what $|\sin z|$ is; after all, you calculate the absolute value using the squares of the real and imaginary parts. So break $z$ into real and imaginary parts, $z = x + iy$. Is there any simplification we can make to $\sin(x+iy)$? Well, if you recall, there’s an addition formula for sine:
$$\sin(A + B) = \sin A\cos B + \cos A\sin B.$$
This is true when $A$ and $B$ are real, but it turns out that it also holds if $A$ and $B$ are complex. (This is a consequence of the principle of permanence of functional equations, one really nice fact of complex analysis.) So we have that
$$\sin(x + iy) = \sin x\cos(iy) + \cos x\sin(iy).$$
At this point we need to figure out what $\cos(iy)$ and $\sin(iy)$ are, and for this, we need to know something about hyperbolic functions. For the rest of this problem, there are two things you need to know about hyperbolic functions:

1. their graph, and
2. two functional equations.

So permit me this brief interlude.

## Hyperbolic Functions

The hyperbolic functions $\cosh$ and $\sinh$ are defined as follows:
$$\cosh t = \frac12(e^t + e^{-t}),\qquad\sinh t = \frac12(e^t-e^{-t}).$$
Here are their graphs (you can disregard $\tanh$):

In particular, cosh has a minimum at $t=0$. When $t$ is large and positive, both $\cosh t$ and $\sinh t$ are very close to $e^t/2$, since the $e^{-t}$ term becomes negligible. In addition, $\cosh$ is even and $\sinh$ is odd; this means that $\cosh(-t)=\cosh t$ and $\sinh(-t)=-\sinh t$.

How are these functions related to $\cos$ and $\sin$? Recall Euler’s formula
$$e^{x+iy} = e^x(\cos y + i\sin y).$$
(Since $\sin(-y)=-\sin y$, it’s also true that $e^{x-iy}=e^x(\cos x – i\sin y)$.)
We see that
$$\cosh it = \frac12(e^{it}+e^{-it}) = \frac12\big((\cos t + i\sin t) + (\cos t – i\sin t)\big) = \cos t$$
and
$$\sinh it = \frac12(e^{it}-e^{-it}) = \frac12\big((\cos t + i\sin t) – (\cos t – i\sin t)\big) = i\sin t.$$
Replacing $t$ with $it$, we see that
$\cos(it) = \cosh(-t) = \cosh t$
and
$i\sin(it) = \sinh(-t)=-\sinh t.$
In summary,
$$\boxed{\cos(it) = \cosh t}\quad\text{ and }\quad \boxed{i\sinh t = \sin(it)}.$$
The boxed equations mean that we can deduce facts about hyperbolic functions from facts about trigonometric functions by replacing $t$ with $it$. For instance, if we take the Pythagorean identity $\cos^2 t + \sin^2 t = 1$ and replace $t$ with $it$, we get that
$$1 = \cos^2(it) + \sin^2(it) = (\cosh t)^2 + (i\sinh t)^2 = \cosh^2 t – \sinh^2 t.$$
This is the second identity you need to know:
$$\boxed{\cosh^2 t – \sinh^2 t = 1}.$$

## The Problem Again

Using the angle addition formula for sine and the functional equations for the hyperbolic functions, we now know that
$$\boxed{\sin(x+iy) = \sin x\cosh y + i\cos x\sinh y}.$$
This is exactly what we wanted — we have separated the real and imaginary parts!

Next, let’s take the absolute value. To avoid writing $\sqrt{}$ over and over, I’m going to consider the square of the absolute value.
$$|\sin(x+iy)|^2 = \sin^2 x\cosh^2 y + \cos^2 x \sinh^2 y.$$
This is pretty good, but it’s still not clear what the size is. We can do a little better by using the Pythagorean identity for trigonometric functions and its analogue for hyperbolic functions. Specifically, using the fact that $\sin^2 x = 1 – \cos^2 x$ gives
$$|\sin(x+iy)|^2 = (1 – \cos^2 x)\cosh^2 y + \cos^2 x \sinh^2 y = \cosh^2 y + \cos^2 x(\sinh^2 y – \cosh^2 y).$$
But we know that $\cosh^2 y – \sinh^2 y = 1$, so in fact
$$\boxed{|\sin(x+iy)|^2 = \cosh^2 y – \cos^2 x}.$$

That was a lot of work, but now we’re basically done. To maximize $|\sin z|$ for $|z|=1$ we want to minimize $\cos x$ and maximize $\cosh y$. Since $\cosh y$ increases in $y$ and $\cos x$ decreases in $x$, we need to maximize $y$ and minimize $x$. The point on the unit circle where this occurs is $\boxed{i}$.

I found this post useful:
Maximum of $\frac{\sin z}{z}$ in the closed unit disc.

By a similar idea of the accepted answer, you can consider the Taylor series of $\sin z$:
$$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\cdots$$

Let $z=i$, all the terms then become a positive number multiplied by $i$. The magnitude is clearly that sum of positive numbers. It then has to be the maximum.

So the maximum happens at $z=i$.