Finding where a complex function is differentiable and holomorphic My function is $f(z)$ = $4$ Re($z$)Im($z$) - $i$($\overline{z}$)$^2$
I simplified my function to be in the form $u(x,y) + iv(x,y)$ then applied the Cauchy Riemann eq's and determined that the function is differentiable on the line $y = -x$ and not holomorphic anywhere since the line has no width.  Is this right?  
Also is there a way to express the derivative of the function in terms of $z$ and not just partials with respect to  $x$ and $y$ ?  My notes from class don't have anything on that.  
 A: To answer your second question, if you know the derivative at a point as a $2 \times 2$ Jacobian matrix of partials, and if you also know from the Cauchy Riemann equations that $f(z)$ is differentiable as a function of $z$ at that point, then that Jacobian matrix must have the form $$\begin{pmatrix} r \cos(\theta) & r \sin(\theta) \\ -r \sin(\theta) & r \cos(\theta) \end{pmatrix}$$ for some $r \ge 0$ and some $\theta$, and therefore you can write $f'(z)$ as the complex number $$f'(z) = r \cos(\theta) + r \sin(\theta) \, i$$
A: Let $z=a+ib$ with real $a,b.$ Then $f(z)=4 a b-i (a-ib)^2=$ $4 ab-i a^2+2 i^2 a b-i^3 b^2=$ $=4 a b-i a^2-2 a b+i b^2=$ $2 a b-i a^2+ i b^2 =$ $(-i)(a^2+2 a b i-b^2)=$ $-i z^2.$
A: To answer your first question:

I simplified my function to be in the form $u(x,y) + iv(x,y)$ then applied the Cauchy Riemann eq's and determined that the function is differentiable on the line $y = -x$ and not holomorphic anywhere since the line has no width.  Is this right? 

This is correct since we require $\mathbb{C}$-differentiability on open subsets of $\mathbb{C}$. The line $y = -x$ contains no open disks/balls of any radius, so $f$ is not holomorphic anywhere.
