How to find where this complex function is differentiable?

Question

This is the complex function:

$$f(z) = 6\bar z^2-2\bar z - 4i |z|^2$$

which is also problem number 7.4, b on page 46 of this book: https://nnquan.files.wordpress.com/2013/01/giao-trinh-ham-phuc.pdf

which I expanded with $z= x+iy$ to get the following:

$$f(z) = (6x^2 + 6y^2-2x) + i(-12xy+2y-4x^2-4y^2)$$

Now, when I apply Cauchy Riemann I get the following equations that must be satisfied:

$$12x-2=-12x+2-8y$$ $$12y = 12y+8x$$

And these are satisfied when $x=0, y=\frac{1}{2}$. Hence, it is differentiable when $x=0, y=\frac{1}{2}$, but not analytic and not entire.

But, the solution on page 49 shows something very different:.

What am I doing wrong?

Answer 1

Something went wrong in your expansion or the differentiation. Cauchy-Riemann is equivalent to saying

$$\left(\frac{\partial f}{\partial \bar{z}}\right)_z = 0$$

and applying we get the expression

$$12\bar{z} - 2 - 4iz = 0 \implies \begin{cases} 3y+x = 0 \\ 6x+2y=1 \\ \end{cases}$$

by separating into real and imaginary components, which does match the solution in your book. Then wherever it is defined, we can find the derivative by saying

$$f'\left(\frac{3}{16}-\frac{1}{16}i\right) = \left(\frac{\partial f}{\partial z}\right)_\bar{z} \Biggr|_{z=\frac{3}{16}-\frac{1}{16}i} = -4i\bar{z} \Biggr|_{z=\frac{3}{16}-\frac{1}{16}i} = \frac{1}{4}-\frac{3}{4}i$$

Answer 2

The first term in $f(z)$ (the real part) is $6x^{2}-6y^{2}-2x$.