Maximum Modulus Principle and Extrema of |sin(z)|

Simplicity is the ultimate sophistication, Anonymous

The Minimum Modulus Principle. Let f be analytic and non-zero in a region A. Then, |f| has no strict local minima on A.

Let’s compare the graph of $|f(z)|$ as a landscape. According to the Maximum Modulus Principle, “mountain peaks” (local maxima) for analytic functions can only appear on the regions’s boundary and never within. The Minimum Modulus Principle says that if the function never touches zero (the “sea level” in our metaphor), then there are no “bottoms of pits” (local minima) inside the region either. The “pits” are also pushed or forced out to the boundary.

Proof.

Since f is analytic and non-zero in a region A, then the reciprocal function $g(z) = \frac{1}{f(z)}$ is well-defined and analytic everywhere in A.

There is an inverse relationship between the magnitude of a number and the magnitude of its reciprocal: $|g(z)| = \left| \frac{1}{f(z)} \right| = \frac{1}{|f(z)|}$. If $z_0$ were a strict local minimum of $|f|$, then $z_0$ would be a strict local maximum of $|g|$.

By the Maximum modulus theorem, $|g(z)| = |\frac{1}{f}|$ cannot have a strict local maximum in A, then $|f(z)|$ cannot have a strict local minimum in A.

Recall: Maximum modulus theorem. A non-constant analytic function cannot have a strict local maximum inside its domain.

Problem. Find the maximum of |sin(z)| on the square domain D = [0, 2π] x [0, 2π].

Solution.

Expand $\sin(z)$ into Real and Imaginary Parts Let $z = x + iy$. $ =$

$$ \begin{aligned} sin(z) = \sin(x+iy) &=[\text{Addition formula for sine}] \sin(x)\cos(iy) + \cos(x)\sin(iy) \\[2pt] &\text{Recall the hyperbolic identities: } \cos(iy) = \cosh(y), \sin(iy) = i\sinh(y)\\[2pt] &= \sin(x)\cosh(y) + i\cos(x)\sinh(y). \end{aligned} $$

sin(x +iy) = sin(x)cosh(y) + isinh(y)cos(x). We want to maximize $|\sin(z)|$. It is mathematically easier to maximize the square, $|\sin(z)|^2$, because the square root function is monotonic (if $A > B$, then $\sqrt{A} > \sqrt{B}$).

$$ \begin{aligned} |\sin(z)|^2 = [\sin(x)\cosh(y)]^2 + [\cos(x)\sinh(y)]^2 &= \sin^2(x)\cosh^2(y) + \cos^2(x)\sinh^2(y) \\[2pt] &\text{Use the identity: } \cosh^2(y) = 1 + \sinh^2(y)\\[2pt] &= \sin^2(x)(1 + \sinh^2(y)) + \cos^2(x)\sinh^2(y) \\[2pt] &\text{Factor out } \sinh^2(y) \\[2pt] &= \sin^2(x) + \sinh^2(y)(\underbrace{\sin^2(x) + \cos^2(x)}_{1}). \end{aligned} $$

Therefore, $|\sin(z)|^2 = \sin^2(x) + \sinh^2(y)$

|sin(x + iy)| = $sin^2(x)cosh^2(y) + sinh^2(y)cos^2(x) = sin^2(x)(1+sinh^2(y)) + sinh^2(y)cos^2(x) = sinh^2(y)(sin^2(x) + cos^2(x)) + sin^2(x) = sinh^2(y) + sin^2(x)$

By the Maximum Modulus Principle, the maximum of a non-constant analytic function on a compact set (like this closed square) must occur on the boundary.

$\sin(z)$ is one of the classical entire functions, along with $e^z, \cos(z)$, polynomials, etc. $\sin z=\sin x\cosh y+i\cos x\sinh y,u(x,y)=\sin x\cosh y, v(x,y)=\cos x\sinh y.$. Its partial derivatives ($u_x=\cos x\cosh y, v_y=\cos x\cosh y, u_y=\sin x\sinh y, v_x=-\sin x\sinh y$) satisfy the Cauchy–Riemann equations ($u_x = v_y, u_y=-v_x$). Since u and v have continuous partial derivatives everywhere, $\sin(z)$ is analytic everywhere.

We need to maximize the sum of two positive terms: $\sin^2(x) + \sinh^2(y)$.

1. The function $\sinh(y)$ is strictly increasing on the entire real line. On our domain $y \in [0, 2\pi]$, the maximum value occurs at the largest $y, y = 2\pi$.

   Recall: $\sinh y=\frac{e^y-e^{-y}}{2}, \frac{d}{dy}\sinh y=\frac{e^y-(-e^{-y})}{2} = \frac{e^y+e^{-y}}{2} \gt 0$

2. The function $\sin^2(x)$ oscillates between 0 and 1. The maximum value is 1, it occurs when $\sin(x) = \pm 1$. On our domain $x \in [0, 2\pi]$, the points where sine is $\pm 1$ are: $x = \frac{\pi}{2} \text{ and } x = \frac{3\pi}{2}$

Since $0 \le sin^2(x) \le 1$ & $sin^2(x) = 1$ where $x = \frac{\pi}{2} \text{ or } \frac{2\pi}{2}$

To get the global maximum, we need both terms to be at their peak simultaneously, $z_1 = \frac{\pi}{2} + i2\pi, z_2 = \frac{3\pi}{2} + i2\pi$.

Hence, the maximum value of $|\sin(z)|$ occurs at the points $\frac{\pi}{2} + 2\pi i$ and $\frac{3\pi}{2} + 2\pi i$. The value of the maximum modulus squared is $1 + \sinh^2(2\pi) = \cosh^2(2\pi)$. Thus, the value of the maximum modulus is $\cosh(2\pi)$.