Proving whether a complex function is analytical help!

Question

I am currently struggling to prove whether a complex function is analytical. I understand that I must employ the Cauchy-Riemann relations to do this. However, the answer I get is one that I can't reconcile with what I am told the answer is. This is the function;

$f(x+iy) = |x^2 - y^2| + 2i|xy|$

With this, I believe that I need to do Cauchy-Riemann with four different functions, namely,

1. $f(x+iy) = x^2 - y^2 + 2ixy$
2. $f(x+iy) = y^2 - x^2 - 2ixy$
3. $f(x+iy) = x^2 - y^2 - 2ixy$
4. $f(x+iy) = y^2 - x^2 + 2ixy$

Which makes me arrive at;

1. $u_x = 2x = v_y = 2x$,

   $u_y = -2y = -v_x = -2y$

2. $u_x = -2x = v_y = -2x$,

   $u_y = 2y = -v_x = 2y$

3. $u_x = 2x = v_y = -2x$,

   $u_y = -2y = -v_x = 2y$

4. $u_x = -2x = v_y = 2x$,

   $u_y = 2y = -v_x = -2y$

respectively.

Now, I think the answer is that the function is only analytical at x = 0, as I think that is the way only all of these equations can be solved simultaneously. But the prescribed solution is not this and I can't really see why.

Thanks :D

Answer 1

You have in the complex plane four open regions where the CR equations are satisfied. These corresponds to yours functions $1$ and $2$ in the regions $U_1=\{ x+iy|x^2-y^2>0\; and\; xy>0 \}$ and $U_2=\{ x+iy|x^2-y^2<0\; and\; xy<0 \}$ respectively.

Edit: Answering your additional questions.

1. " functions 3 and 4 do not satisfy CR (not even at 0?)"

For analicity in a point $z_0$ we need CR equations be satisfied on open set that contains $z_0$. As no such a set exist for $z_0=0$, your function is not analitic there.

2. "but they do correspond to the initial function, right? Like I expanded the absolute values correctly? "

Yes. All yous considerations where correct.

3. "You're also saying functions 1 and 2 satisfy CR in those certain regions, I still can't see how to arrive at that."

It's just your own calculations. For cases 1 and 2, where the absolute values are the identity or minus identity, the equations are satisfied. In cases 3 and 4 no. Note that case 1 is: $|x^2-y^2|=x^2-y^2$ and $|xy|=xy$, that is: $x^2-y^2>0$ and $xy>0$. Also case 2 is: $|x^2-y^2|=-(x^2-y^2)$ and $|xy|=-(xy)$, that is: $x^2-y^2<0$ and $xy<0$.