# How to show if a complex function is analytic?

Just began the study of complex analysis. Let $$f(x,y) = x^2 - y^2 + 2 i xy - x - iy.$$ I need to determine if this function is analytic. This means I have to show the partials satisfy the Cauchy-Riemann equations, and that the partials are continuous. So in this case we have $u(x,y) = x^2 - y^2 - x$ and $v(x,y) = 2xy - y$. Now \begin{align*} \frac{\partial v}{\partial y} = 2x - 1 = \frac{\partial u}{\partial x} \end{align*} and \begin{align*} - \frac{\partial u}{\partial y} = - (-2y) = 2 y = \frac{\partial v}{\partial x} \end{align*} Hence the Cauchy-Riemann equations are satisfied, which is a necessary condition for being analytic, but not sufficient. Now I have to show the partials are continuous? How do I do that?

• Get out your epsilons and deltas, in the usual way? – Chappers Sep 26 '15 at 15:19
• In a complex analysis course it's safe to assume you can just say that the partials are continuous because they are polynomials. – Git Gud Sep 26 '15 at 15:20
• No need to show that the partial derivatives are continuous. CR + continuity of $f$ is enough. – A.Γ. Sep 26 '15 at 17:26

## 3 Answers

Let $g$ define by $g(z)=z^2-z$. $g$ is clearly analytic on $\mathbb C$. You have that $f(x,y)=g(z(x,y))$ which is a composition of two analytic function.

We have that $u = u(x,y)$ and $v=v(x,y)$ are polynomial functions of $x$ and $y$. If you know that polynomials are continuous functions and that derivatives of polynomials are again polinomials, you're done.

Another way is to transform your function into a function of $$z^{-}$$, where z^ is the conjugate of z, i.e., if: $$z= x+iy , z^{-}=x -iy$$:

$$x= \frac {z+z^{-}}{2}$$ $$y= \frac { z -z^{-}}{2i}$$

and the resulting function should be a function of $$z$$ exclusively. The simplest example likely is : $$f(z)= x+iy$$ Then : $$f(z)= \frac {z+z^{-}}{2}+ i\frac{z-z^{-}}{2i}= \frac {z+z^{-}}{2}+ \frac{z-z^{-}}{2}=z$$