# 7.4 The maximum modulus principle

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Our next goal is to examine so-called “max/min” problems for coplex-valued functions of complex variables.Since order makes no sense for complex numbers, we will investigate max/min problems for the absolute value of a complex-valued function.For the corresponding question for real-valued functions of real variables, we have as our basic resultthe First Derivative Test. Indeed, when searching for the poinhts where a differentiable real-valued function $f$ on an interval $\left[a,b\right]$ attains its extreme values, we consider first the poinhts where it attains a local max or min.Of course, to find the absolute minimum and maximum, we must also check the values of the function at the endpoints.

Our next goal is to examine so-called “max/min” problems for coplex-valued functions of complex variables.Since order makes no sense for complex numbers, we will investigate max/min problems for the absolute value of a complex-valued function.For the corresponding question for real-valued functions of real variables, we have as our basic resultthe First Derivative Test ( [link] ). Indeed, when searching for the poinhts where a differentiable real-valued function $f$ on an interval $\left[a,b\right]$ attains its extreme values, we consider first the poinhts where it attains a local max or min, to which purpose end [link] is useful. Of course, to find the absolute minimum and maximum, we must also check the values of the function at the endpoints.

An analog of [link] holds in the complex case, but in fact a much different result is really valid. Indeed, it is nearly impossible for the absolute value ofa differentiable function of a complex variable to attain a local maximum or minimum.

Let $f$ be a continuous function on a piecewise smooth geometric set $S,$ and assume that $f$ is differentiable on the interior ${S}^{0}$ of $S.$ Suppose $c$ is a point in ${S}^{0}$ at which the real-valued function $|f|$ attains a local maximum. That is, there exists an $ϵ>0$ such that $|f\left(c\right)|\ge |f\left(z\right)|$ for all $z$ satisfying $|z-c|<ϵ.$ Then $f$ is a constant function on $S;$ i.e., $f\left(z\right)=f\left(c\right)$ for all $z\in S.$ In other words, the only differentiable functions of a complex variable, whose absolute value attains a local maximum on the interior of a geometric set, are constant functions on that set.

If $f\left(c\right)=0,$ then $f\left(z\right)=0$ for all $z\in {B}_{ϵ}\left(c\right).$ Hence, by the Identity Theorem ( [link] ), $f\left(z\right)$ would equal 0 for all $z\in S.$ so, we may as well assume that $f\left(c\right)\ne 0.$ Let $r$ be any positive number for which the closed disk ${\overline{B}}_{r}\left(c\right)$ is contained in ${B}_{ϵ}\left(c\right).$ We claim first that there exists a point $z$ on the boundary ${C}_{r}$ of the disk ${\overline{B}}_{r}\left(c\right)$ for which $|f\left(z\right)|=|f\left(c\right)|.$ Of course, $|f\left(z|\le |f\left(c\right)|$ for all $z$ on this boundary by assumption. By way of contradiction, suppose that $|f\left(\zeta \right)|<|f\left(c\right)|$ for all $\zeta$ on the boundary ${C}_{r}$ of the disk. Write $M$ for the maximum value of the function $|f|$ on the compact set ${C}_{r}.$ Then, by our assumption, $M<|f\left(c\right)|.$ Now, we use the Cauchy Integral Formula:

$\begin{array}{ccc}\hfill |f\left(c\right)|& =& |\frac{1}{2\pi i}{\int }_{{C}_{r}}\frac{f\left(\zeta \right)}{\zeta -c}\phantom{\rule{0.166667em}{0ex}}d\zeta |\hfill \\ & =& \frac{1}{2\pi }|{\int }_{0}^{2\pi }\frac{f\left(c+r{e}^{it}\right)}{r{e}^{it}}ir{e}^{it}\phantom{\rule{0.166667em}{0ex}}dt|\hfill \\ & \le & \frac{1}{2\pi }{\int }_{0}^{2\pi }|f\left(c+r{e}^{it}\right)|\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & \le & \frac{1}{2\pi }{\int }_{0}^{2\pi }M\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & =& M\hfill \\ & <& |f\left(c\right)|,\hfill \end{array}$

and this is a contradiction.

Now for each natural number $n$ for which $1/n<ϵ,$ let ${z}_{n}$ be a point for which $|{z}_{n}-c|=1/n$ and $|f\left({z}_{n}\right)|=|f\left(c\right)|.$ We claim that the derivative ${f}^{\text{'}}\left({z}_{n}\right)$ of $f$ at ${z}_{n}=0$ for all $n.$ What we know is that the real-valued function $F\left(x,y\right)={|f\left(x+iy\right)|}^{2}{=\left(u\left(x,y\right)}^{2}+{\left(v\left(x,y\right)\right)}^{2}$ attains a local maximum value at ${z}_{n}=\left({x}_{n},{y}_{n}\right).$ Hence, by [link] , both partial derivatives of $F$ must be 0 at $\left({x}_{n},{y}_{n}\right).$ That is

$2u\left({x}_{n},{y}_{n}\right)\frac{tialu}{tialx}\left({x}_{n},{y}_{n}\right)+2v\left({x}_{n},{y}_{n}\right)\frac{tialv}{tialx}\left({x}_{n},{y}_{n}\right)=0$

and

$2u\left({x}_{n},{y}_{n}\right)\frac{tialu}{tialy}\left({x}_{n},{y}_{n}\right)+2v\left({x}_{n},{y}_{n}\right)\frac{tialv}{tialy}\left({x}_{n},{y}_{n}\right)=0.$

Hence the two vectors

${\stackrel{\to }{V}}_{1}=\left(\frac{tialu}{tialx}\left({x}_{n},{y}_{n}\right),\frac{tialv}{tialx}\left({x}_{n},{y}_{n}\right)\right)$

and

${\stackrel{\to }{V}}_{2}=\left(\frac{tialu}{tialy}\left({x}_{n},{y}_{n}\right),\frac{tialv}{tialy}\left({x}_{n},{y}_{n}\right)\right)$

are both perpendicular to the vector ${\stackrel{\to }{V}}_{3}=\left(u\left({x}_{n},{y}_{n}\right),v\left({x}_{n},{y}_{n}\right)\right).$ But ${\stackrel{\to }{V}}_{3}\ne 0,$ because $\parallel {\stackrel{\to }{V}}_{3}\parallel =|f\left({z}_{n}\right)|=|f\left(c\right)|>0,$ and hence ${\stackrel{\to }{V}}_{1}$ and ${\stackrel{\to }{V}}_{2}$ are linearly dependent. But this implies that ${f}^{\text{'}}\left({z}_{n}\right)=0,$ according to [link] .

Since $c=lim{z}_{n},$ and ${f}^{\text{'}}$ is analytic on ${S}^{0},$ it follows from the Identity Theorem that there exists an $r>0$ such that ${f}^{\text{'}}\left(z\right)=0$ for all $z\in {B}_{r}\left(c\right).$ But this implies that $f$ is a constant $f\left(z\right)=f\left(c\right)$ for all $z\in {B}_{r}\left(c\right).$ And thenm, again using the Identity Theorem, this implies that $f\left(z\right)=f\left(c\right)$ for all $z\in S,$ which completes the proof.

REMARK Of course, the preceding proof contains in it the verification that if $|f|$ attains a maximum at a point $c$ where it is differentiable, then ${f}^{\text{'}}\left(c\right)=0.$ This is the analog for functions of a complex variable of [link] . But, [link] certainly asserts a lot more than that. In fact, it says that it is impossible for the absolute value of a nonconstant differentiable function of a complexvariable to attain a local maximum. Here is the coup d'grâs:

## Maximum modulus principle

Let $f$ be a continuous, nonconstant, complex-valued function on a piecewise smooth geometric set $S,$ and suppose that $f$ is differentiable on the interior ${S}^{0}$ of $S.$ Let $M$ be the maximum value of the continuous, real-valued function $|f|$ on $S,$ and let $z$ be a point in $S$ for which $|f\left(z\right)|=M.$ Then, $z$ does not belong to the interior ${S}^{0}$ of $S;$ it belongs to the boundary of $S.$ In other words, $|f|$ attains its maximum value only on the boundary of $S.$

1. Prove the preceding corollary.
2. Let $f$ be an analytic function on an open set $U,$ and let $c\in U$ be a point at which $|f|$ achieves a local minimum; i.e., there exists an $ϵ>0$ such that $|f\left(c\right)|\le |f\left(z\right)|$ for all $z\in {B}_{ϵ}\left(c\right).$ Show that, if $f\left(c\right)\ne 0,$ then $f$ is constant on ${B}_{ϵ}\left(c\right).$ Show by example that, if $f\left(c\right)=0,$ then $f$ need not be a constant on ${B}_{ϵ}\left(c\right).$
3. Prove the “Minimum Modulus Principle:” Let $f$ be a nonzero, continuous, nonconstant, function on a piecewise smooth geometric set $S,$ and let $m$ be the minimum value of the function $|f|$ on $S.$ If $z$ is a point of $S$ at which this minimum value is atgtained, then $z$ belongs to the boundary ${C}_{S}$ of $S.$

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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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