Complex AnalysisContour Integrals and Cauchy's Theorem

The Maximum Modulus Principle

On a bounded region, a nonconstant analytic function that is continuous up to the boundary reaches its largest |f| on the boundary, never inside.

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Take Cauchy's formula on a circle centered at $a$: it says $f(a)$ is the average of $f$ over the circle. An average can never be bigger than every value it averages, and that simple observation controls where $|f|$ can be largest.

Key idea

Maximum modulus principle. If $f$ is analytic and not constant on a bounded region $D$ and continuous on its closure, then $|f|$ reaches its maximum on the boundary of $D$, never at an interior point.
If moreover $f$ has no zeros in $D$, the minimum of $|f|$ is on the boundary too (apply the principle to $\frac{1}{f}$).

Why it works

With $z = a + re^{it}$ in Cauchy's formula, $dz = ire^{it}dt$ and $z - a = re^{it}$, so $f(a) = \frac{1}{2\pi}\int_0^{2\pi} f(a + re^{it})\,dt$: the mean value property. Suppose $|f|$ had its largest value $M$ at an interior point $a$. Then $M = |f(a)| \le \frac{1}{2\pi}\int |f(a + re^{it})|\,dt \le M$, so equality holds and $|f| = M$ on every small circle around $a$. Then $|f|$ is constant near $a$, which (by the Cauchy–Riemann equations) forces $f$ to be constant.
$f(z) = z^2 + z + 1$ on the closed unit disk: the largest $|f|$ is marked, on the edge at $z = 1$ where $|f| = 3$. Inside the disk the brightness rings never reach that level.

Worked examples

Worked example: A linear function

Find the maximum of $|z + 3 - 4i|$ on $|z| \le 2$ and where it happens.
  1. The maximum is on the circle $|z| = 2$.
  2. $|z + (3 - 4i)| \le |z| + |3 - 4i| = 2 + 5 = 7$, with equality when $z$ points the same way as $3 - 4i$.
  3. $z = 2\cdot\frac{3 - 4i}{5} = \frac{6}{5} - \frac{8}{5}i$, where the value is $7$.

Answer: $7$, at $\frac{6}{5} - \frac{8}{5}i$

Worked example: An exponential

Find the maximum of $|e^{iz}|$ on $|z| \le 1$.
  1. $|e^{iz}| = e^{\operatorname{Re}(iz)} = e^{-y}$.
  2. On the closed disk, $-y$ is largest at $z = -i$, where $-y = 1$.
  3. The maximum is $e$, at $z = -i$, on the boundary.

Answer: $e$, at $-i$

Worked example: A minimum

Find the minimum of $|z + 3|$ on $|z| \le 1$.
  1. $z + 3$ has its only zero at $-3$, outside the disk, so the minimum is on the boundary.
  2. $|z + 3| \ge 3 - |z| = 2$, reached at $z = -1$.

Answer: $2$

The same idea works for the real part: a harmonic function $u$ (such as $\operatorname{Re} f$) takes its maximum and minimum over a closed bounded region on the boundary. In heat flow this says the steady temperature inside a plate is never hotter than the hottest point of its edge.

Common mistake

Hunting for the maximum at a critical point inside. In multivariable calculus you set the gradient to zero and test interior points. For $|f|$ with $f$ analytic, interior critical points are never maxima (and $f'(z_0) = 0$ is a different thing altogether). Go straight to the boundary: parametrize it, or use the triangle inequality and check when equality holds.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Builds on

Leads to

This is one of the top rungs of Complex Analysis.

Sources and further reading

  • Beck, Marchesi, Pixton and Sabalka, A First Course in Complex Analysis, chapter 5 (Consequences of Cauchy's Theorem)
  • Beck, Marchesi, Pixton and Sabalka, A First Course in Complex Analysis, chapter 6 (Harmonic Functions)
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