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How do you find the maximum of the complex function $|\cos{z}|$ on $[0,2\pi]\times[0,2\pi]$. I believe I'm to use the maximum modulus principle, since the function is entire. I'm just having problems starting. Any suggestions?

L.A Rhoads
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  • Start by considering the values of $\cos z$ on the bottom sides of the square. Do you know what they are? –  May 06 '15 at 01:01
  • The bottom corners of the square should be both 1, correct? |cos(0+0i)|=1, and |cos(2pi+0i)|=1 as well. – L.A Rhoads May 06 '15 at 17:27
  • On the interval from $0$ to $2\pi$ the function $\cos x$ attains all values between $-1$ and $1$. Therefore, the maximum of $|\cos x|$ there is $1$. On other sides, use the relation with complex exponential. –  May 06 '15 at 17:34

2 Answers2

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Because cos z is entire, |cos z| has its maximum along the boundary of the square [0,2$\pi$]$\times$[0,2$\pi$], by the Maximum Modulus Principle.

Let cos z = cos($x+iy$), so that we can check its values at the boundaries. Although cos $\theta$ for real values just oscillates between 1 and -1, we are not dealing with real values.

It just so happens that |cos($\tilde{x}+2\pi i$)|=267.746761483748... (a transcendental number) for $\tilde{x}=\frac{n\pi}{2}$ for n=0,1,2,3,4. You can show this using the exponential representation of cos($x+iy$). Thus, the maximum is 267.746761483748...

Morgan Henderson
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The idea is to apply the same technique used in one of the answers of this problem Maximum of $|\sin(z)|$ as $\{z: |z| \leq 1 \} $

Use the identity: $$\vert\cos z\vert^2=\vert \cos (x+iy)\vert^2=\cos^2 x + \sinh ^2 y$$.

What is left is just to find the maximum of $\cos x$ and the maximum of $\sinh y$ in the boundary. Since $x, y \in[0,2\pi]$, the maximum of $\cos x$ is $1$ when $x=0,\,2\pi$ and the maximum of $\sinh$ is $267.74489404102$ when $y=2\pi$.

Finally by the Maximum Modulus Principle $$\max_{[0,2\pi]\times[0,2\pi]}\vert \cos z\vert=\sqrt{1^2+267.74489404102^2}\approx 267.74676148375$$

Luis E. Mack
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