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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 ( [link] ). Indeed, when searching for the poinhts where a differentiable real-valued function $f$ on an interval $[a,b]$ 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 $\left|f\right|$ attains a local maximum. That is, there exists an $\u03f5>0$ such that $\left|f\right(c\left)\right|\ge \left|f\right(z\left)\right|$ for all $z$ satisfying $|z-c|<\u03f5.$ 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 $z\in {B}_{\u03f5}\left(c\right).$ 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 ${B}_{\u03f5}\left(c\right).$ 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 $\left|f\right(z\left)\right|=\left|f\right(c\left)\right|.$ Of course, $\left|f\right(z|\le |f\left(c\right)|$ for all $z$ on this boundary by assumption. By way of contradiction, suppose that $\left|f\right(\zeta \left)\right|<\left|f\right(c\left)\right|$ for all $\zeta $ on the boundary ${C}_{r}$ of the disk. Write $M$ for the maximum value of the function $\left|f\right|$ on the compact set ${C}_{r}.$ Then, by our assumption, $M<\left|f\right(c\left)\right|.$ Now, we use the Cauchy Integral Formula:
and this is a contradiction.
Now for each natural number $n$ for which $1/n<\u03f5,$ 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 $F(x,y)={\left|f(x+iy)\right|}^{2}{=(u(x,y)}^{2}+{\left(v(x,y)\right)}^{2}$ attains a local maximum value at ${z}_{n}=({x}_{n},{y}_{n}).$ Hence, by [link] , both partial derivatives of $F$ must be 0 at $({x}_{n},{y}_{n}).$ That is
and
Hence the two vectors
and
are both perpendicular to the vector ${\overrightarrow{V}}_{3}=(u({x}_{n},{y}_{n}),v({x}_{n},{y}_{n})).$ But ${\overrightarrow{V}}_{3}\ne 0,$ because $\parallel {\overrightarrow{V}}_{3}\parallel =|f\left({z}_{n}\right)|=|f\left(c\right)|>0,$ and hence ${\overrightarrow{V}}_{1}$ and ${\overrightarrow{V}}_{2}$ 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 $\left|f\right|$ 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:
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 $\left|f\right|$ on $S,$ and let $z$ be a point in $S$ for which $\left|f\right(z\left)\right|=M.$ Then, $z$ does not belong to the interior ${S}^{0}$ of $S;$ it belongs to the boundary of $S.$ In other words, $\left|f\right|$ attains its maximum value only on the boundary of $S.$
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