Cauchy Riemann equations necessary and sufficient condition?

Question

I was taught that $f(z)$ is differentiable at $z_0=x_0+y_0$ iff Cauchy Riemann equations hold at $(x_0,y_0)$. However, I was shown this example: $f(z)=\frac{\operatorname{Re}(z) \cdot \operatorname{Im}(z)}{z}, z \neq 0$ and $f(0)=0$. So Cauchy-Riemann equations hold at $(0,0)$, however the function is not differentiable at $(0,0)$.

So is it indeed true that $f(z)$ is differentiable at $z_0=x_0+y_0$ iff Cauchy Riemann equations hold at $(x_0,y_0)$?

Answer 1

$f(z)$ is complex-differentiable if and only if:

1. It is differentiable as a function $\Bbb{R}^2\to\Bbb{R}^2$, for example, if $F(x,y):=Re[f(x+iy)]$ and $G(x,y):=Im[f(x+iy)]$, then: $$ \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y},\frac{\partial G}{\partial x}, \frac{\partial G}{\partial y} $$ exist (see Daniel Fischer's comment), and:
2. The Cauchy-Riemann conditions hold, i.e. in the notation above: $$ \frac{\partial F}{\partial x}= \frac{\partial G}{\partial y},\frac{\partial G}{\partial x}=- \frac{\partial f}{\partial y}. $$

Note that for the second to hold, the first one has to hold, otherwise the expressions do not make sense.

In the example you gave, the function is not differentiable in $(0,0)$.

Answer 2

Try to show that $f(z)$ doesn't tend to a unique limit as $(x,y)\to (0,0)$ along any curve through origin.

You can choose $y= mx$ such that

$f(x,mx) = \frac{m}{(1+im)^2}$ as $x\to 0$ which depends on m. So, the function fails to exist uniquely at $z=0$ which shows that even the $lim_{z\rightarrow 0} f(z)$ doesn't exist.