Maxima and Minima of a complex function

**sashikanth** · April 11th 2012, 12:13 PM

Let $f(z) = z^2 - z$ in the circular region $R = {z: |z| \leq 1}$ . Find points in $R$ where $|f(z)|$ has its maximum or minimum values.

An attempt at a solution: Use $z = x+iy$ to convert the given function into one in x and y, and then take the norm using the standard formula and differentiate partially wrt x, y to get extrema points.

The problem is that the resulting function of x, y is too cumbersome and when I differentiate and try to solve the 2 resulting equations, I am unable to solve it. Is there any other approach to solve such problems?

**Prove It** · April 11th 2012, 04:51 PM

Originally Posted by sashikanth

Let $f(z) = z^2 - z$ in the circular region $R = {z: |z| \leq 1}$ . Find points in $R$ where $|f(z)|$ has its maximum or minimum values.

An attempt at a solution: Use $z = x+iy$ to convert the given function into one in x and y, and then take the norm using the standard formula and differentiate partially wrt x, y to get extrema points.

The problem is that the resulting function of x, y is too cumbersome and when I differentiate and try to solve the 2 resulting equations, I am unable to solve it. Is there any other approach to solve such problems?

First, write $\displaystyle \begin{align*} z = x + i\,y \end{align*}$ , so that

$\displaystyle \begin{align*} f(z) &= z^2 - z \\ f(x + i\,y) &= (x + i\,y)^2 - (x + i\,y) \\ &= x^2 - y^2 + 2i\,x\,y - x - i\,y \\ &= x^2 - x - y^2 + i\left(2x\,y - y\right) \\ &= u(x, y) + i\,v(x, y) \end{align*}$

Now we need to check the Cauchy-Riemann Equations hold so that we know we can differentiate the function: $\displaystyle \begin{align*} \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \end{align*}$ and $\displaystyle \begin{align*} \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \end{align*}$ .

$\displaystyle \begin{align*} \frac{\partial u}{\partial x} &= 2x - 1 \\ \\ \frac{\partial v}{\partial y} &= 2x - 1 \\ &= \frac{\partial u}{\partial x} \textrm{ as required} \\ \\ \frac{\partial u}{\partial y} &= -2y \\ \\ \frac{\partial v}{\partial x} &= 2y \\ &= -\frac{\partial u}{\partial y} \textrm{ as required} \end{align*}$

So the Cauchy Riemann Equations hold for all x and y, which means the complex function can be differentiated everywhere. We know that maxima and minima occur where the derivative is 0, so

$\displaystyle \begin{align*} f(z) &= z^2 - z \\ f'(z) &= 2z - 1 \\ 0 &= 2z - 1 \\ z &= \frac{1}{2} \\ \\ f''(z) &= 2 \end{align*}$

Since the second derivative is positive, the function is minimised when $\displaystyle \begin{align*} z = \frac{1}{2} \end{align*}$ and the minimum value is $\displaystyle \begin{align*} f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = -\frac{1}{4} \end{align*}$ .

**sashikanth** · April 11th 2012, 07:13 PM

Hi, thanks for the reply! The questions asks the maximum and minimum values of $|f(z)|$ , so it would be the extrema of $\sqrt{(u(x,y)^2 + v(x,y)^2)}$ . Thats the part which becomes cumbersome when you deal with x and y

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