Is the function $$f(z)=|z^{2}-z|$$ nowhere analytic? Justify your ansewr
WHat i tried
Let $z=x+iy$ and then substituting it to the function above to get the form $$f(x+iy)=u(x,y)+i(x,y)$$ where $u(x,y)$ is the real part and $v(x,y)$ the
imaginary part and then using the Cauchy Riemann equations to show that it holds.ie
$$U_{x}=V_{y}$$
$$U_{y}=-V_{x}$$
Then find the four partial derivatives $U_{x}$, $U_{y}$ , $V_{x}$, $V_{y}$ and if they are continuous then the function are analytic. My problem is that because of the modulus i couldent differentiate and find the partial derivatives in the usual way. Do i have to split up the modulus into the nehgative and positive portion and find the partial derivatives for each case. Could anyone explain. Thanks
Best Answer
Hint: The modulus of a complex number $a+ib$ is $\sqrt{a^2+b^2}$. Use this to identify $u$ and $v$ as functions of $x,y$. Hint2: There is one point at which $f$ is complex differentiable.