# 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? Sep 26, 2015 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. Sep 26, 2015 at 15:20
• No need to show that the partial derivatives are continuous. CR + continuity of $f$ is enough.
– A.Γ.
Sep 26, 2015 at 17:26

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 $$\overline{z}$$, where $$\overline{z}$$ is the conjugate of $$z$$, i.e.,

$$z= x + iy ,\quad \overline{z} = x -iy.$$

In terms of $$z$$ and $$\overline{z}$$, we can write $$x$$ and $$y$$ as

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

The given function is written in terms of $$z$$ and $$\overline{z}$$. An analytic function should be a function of $$z$$ exclusively.

The simplest example likely is \begin{align} f(z) &= x+iy \\ &= \frac {z+\overline{z}}{2}+ i\frac{z-\overline{z}}{2i} \\ &= \frac {z+\overline{z}}{2}+ \frac{z-\overline{z}}{2} \\ &= z, \end{align} which means $$f$$ is analytic.