So I've been asked
'Is $f$ analytic anywhere? Everywhere? Justify your answer.'
My function is $f(z)=z^2-iz+iz^2$ which I've expressed as $f=u+iv$ so I can use Cauchy-Riemann equations to test to see whether or not $f$ is analytic.
I've shown the Cauchy-Riemann equations do indeed hold. I have
$$u_x=2x-2y=v_y \\ v_x=2x+2y+1=-u_y$$
… I don't understand how to determine if it's anywhere or everywhere… I believe it's everywhere, but how could I tell?
Best Answer
Note that "The sole existence of partial derivatives satisfying the Cauchy–Riemann equations is not enough to ensure complex differentiability at that point. It is necessary that u and v be real differentiable, which is a stronger condition than the existence of the partial derivatives, but it is not necessary that these partial derivatives be continuous." (from wiki).
However, in your case Cauchy–Riemann equations hold for all $(x,y)$ and $u$ and $v$ are everywhere real differentiable, so $f$ is analytic everywhere.
In general, if it's satisfied at some point $(x,y)$ then $f$ is differentiable at this point.