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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 $(x_0,y_0)$?

RobPratt
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Hjersh
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3 Answers3

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$f(z)$ 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 $(0,0)$.

geodude
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    The continuity of the partial derivatives is stronger than real differentiability. – Daniel Fischer Jun 24 '15 at 13:06
  • Right...which one of the two criteria is then used for complex differentiability? – geodude Jun 24 '15 at 13:08
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    Real differentiability + Cauchy-Riemann, where real differentiability (Fréchet differentiability) means that $$\lim_{z\to z_0} \frac{f(z) - f(z_0) - \frac{\partial f}{\partial x}(z_0)(x-x_0) - \frac{\partial f}{\partial y}(z_0)(y-y_0)}{z-z_0} = 0.$$ In the example, the CR equations hold only at $0$ - which is the only point where the function is not real differentiable. – Daniel Fischer Jun 24 '15 at 13:12
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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.

Nitin Uniyal
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Having a complex derivative at a point $z$, $$ \lim_{h\to0}\frac{f(z+h)-f(z)}{h}=f'(z), $$ or for comparision, the existence of a number $f'(z)=a+ib$ $$ \lim_{h\to0}\frac{f(z+h)-f(z)-hf'(z)}{h}=0, $$ is equivalent to being real differentiable at $z$ (abusing notation slightly for convenience, $\mathbb{R}^2\cong\mathbb{C}$ as real vector spaces) $$ \lim_{h\to 0}\frac{|f(z+h)-f(z)-L(h_1,h_2)|}{|h|}=0 $$ where the linear map $L$ is the matrix of multiplication by a complex number (the derivative of $f$ at $z$) $$ L(h_1,h_2)= \left( \begin{array}{cc} a&b\\ -b&a\\ \end{array} \right) \left( \begin{array}{c} h_1\\ h_2\\ \end{array} \right)\sim (a+ib)(h_1+ih_2), \quad f'(z)=a+ib. $$ The symmetries in the matrix for $L$ are the Cauchy--Riemann equations.

yoyo
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