# Complex continuity and differentiability of a piecewise function

Let $g(z):\mathbb{C}\rightarrow\mathbb{C}$ with $$g(z)=g(x+iy) = \begin{cases} \dfrac{x^2y+ixy^2}{x^2+y^2} & : (x,y)\neq (0,0), \\ 0 & : (x,y)=(0,0). \end{cases}$$

1 How do I show that $g$ is continuous at $z=0$?
2 How do I show that $g$ is not complex differentiable at $z=0$?

What I know:
1 I know that the composition of continuous functions is continuous, but I don't know how to show that the 'subfunctions' of $g$ are continuous. How would I do this with the $\varepsilon$-$\delta$-definition of continuity?
2 I know that $g$ satisfies the Cauchy-Riemann equations at $z=0$, but this is not enough right? My textbook says that $g$ must be also totally differentiable at $z=0$ in the sense of real analysis ($\mathbb{C}=\mathbb{R}^2$). How would I show this?

Edit: $g$ is NOT differentiable at $z=0$.

• To show continuity at $0$, it may be helpful to think in polar coordinates. Regarding point 2, you should rather try to show that $g$ is not complex differentiable at $0$. – Daniel Fischer Mar 6 '16 at 16:14
• @DanielFischer My question says to show that it is compl. differentiable – Spider-Pig Mar 6 '16 at 16:28
• Also how would the polar coordinates help? – Spider-Pig Mar 6 '16 at 16:28
• It may be easier to see continuity at $0$ if you think of polar coordinates. It may not be. The question may ask to show that it's complex differentiable at $0$, but it isn't. There's a mistake in the question then. – Daniel Fischer Mar 6 '16 at 16:41
• @DanielFischer Could you maybe show why it isn't complex differentiable at $0$? – Spider-Pig Mar 6 '16 at 17:18

Outside of the origin, it is convenient to write $g$ in a different manner. The real and imaginary part of the numerator have the common factor $xy$, and taking that out, we can write

$$g(x+iy) = \begin{cases} \dfrac{xy}{x^2+y^2}(x+iy) &, (x,y) \neq (0,0) \\ \qquad 0 &, (x,y) = (0,0). \end{cases}\tag{1}$$

In that form, the answer to both questions is easily found by looking at the factor $\frac{xy}{x^2+y^2}$. The inequality between arithmetic and geometric mean shows

$$\lvert xy\rvert \leqslant \frac{x^2+y^2}{2},$$

and hence $\Bigl\lvert \frac{xy}{x^2+y^2}\Bigr\rvert \leqslant \frac{1}{2}$. This shows

$$\lvert g(x+iy)\rvert \leqslant \frac{1}{2}\lvert x+iy\rvert$$

for $x+iy \neq 0$, from which the continuity of $g$ at $0$ follows easily.

For the question of complex differentiability, $(1)$ lets us write the difference quotient

$$\frac{g(x+iy) - g(0)}{x+iy} = \frac{xy}{x^2+y^2},\tag{2}$$

and one notes that the right hand side has no limit as $(x,y) \to (0,0)$, every value in $\bigl[-\frac{1}{2},\frac{1}{2}\bigr]$ is attained on every circle $\{x+iy : \sqrt{x^2+y^2} = \rho\}$. On the ray $\{ r(\cos \varphi + i\sin\varphi) : r \in (0,+\infty)\}$ the difference quotient has the constant value $\frac{1}{2}\sin (2\varphi)$. Thus $g$ is not complex differentiable at $0$.

This function $g$ shows that it is not sufficient to imply complex differentiability that the partial derivatives exist and satisfy the Cauchy-Riemann equations, (total) real differentiability (Fréchet differentiability) is required.