Apparently, it can be shown that the Cauchy-Riemann equations can be written simply as, $df/dz^*=0$. I do not understand how it does not immediately follow from this that $df/dz=0$.
When we proved the relations originally, we used $$\frac{df}{dz} = \frac{\delta u+i\delta v}{\delta x+i\delta y}$$ Taking both the limits $\delta x\to0$ and $\delta y \to 0$, and requiring they be equal for the derivative to be defined.
Doing the same thing for $df/dz^*$, we get exactly the same thing for $\delta x\to 0$. Since this has to be zero, haven't we also shown that $df/dz=0$ if $df/dz^*$ is defined? Or am I missing something obvious?
Thanks!
$\endgroup$ 12 Answers
$\begingroup$The derivative is defined as $$ f'(z) = \lim_{\Delta z\to 0}\frac{f(z+\Delta z) - f(z)}{\Delta z}. $$ Let $f(z) = u(x,y) + iv(x,y)$. The limit must be the same no matter how we set up $\Delta z\to 0$. Suppose we choose only real values for $\Delta z$. The imaginary part is a constant so the derivative is with respect to $x$. $$ f'(z) = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} $$ Now if we hold the reals constant, we have $$ f'(z) = -i\frac{\partial f}{\partial y} = -i\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y} $$ since we would have had $$ f'(z) = \lim_{\Delta z\to 0}\frac{f(z+i\Delta z) - f(z)}{i\Delta z}. $$ Therefore, $f_x = -if_y$ or $$ u_x = v_y\qquad\text{and}\qquad u_y = -v_x $$ which are the Cauchy Riemann equations. Let $f(x,y)$ be a complex functions of real variables $x,y$. Let $z=x+iy$ and $\bar{z}=x-iy$ so $$ x=(z+\bar{z})/2\qquad\text{and}\qquad y = -i(z-\bar{z})/2. $$ Then $$ \frac{\partial f}{\partial z} = 1/2\Bigl[\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\Bigr]\qquad\text{and}\qquad \frac{\partial f}{\partial\bar{z}} = 1/2\Bigl[\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\Bigr] $$ Recall that $f_x = -if_y$ so $$ \frac{\partial f}{\partial\bar{z}} = 1/2\Bigl[\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\Bigr] = 1/2\Bigl[-i\frac{\partial f}{\partial y}+i\frac{\partial f}{\partial y}\Bigr] = 0. $$ Since $f_x = -if_y$, $$ \frac{\partial f}{\partial z} = 1/2\Bigl[\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\Bigr] = 1/2\Bigl[-i\frac{\partial f}{\partial y}-i\frac{\partial f}{\partial y}\Bigr] = -i\frac{\partial f}{\partial y}. $$
$\endgroup$ 5 $\begingroup$The derivation given in the other answer is purely formal, as Ahlfors points out:(The equation (5) is just $$\frac{\partial f}{\partial x}=\frac{1}{i}\frac{\partial f}{\partial y},$$ which is of course equivalent to the usual CR equations.) Notice how he says "if the rules of calculus were applicable"; these manipulations are purely symbolic, and one must define $\frac{\partial f}{\partial\bar{z}}$ by the equation$$\frac{\partial f}{\partial\bar{z}}:=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\frac{1}{i}\frac{\partial f}{\partial y} \right) .$$From this is clear that the equation $$\frac{\partial f}{\partial\bar{z}}=0$$ is equivalent to the usual CR equations. One cannot really derive this equation (as it is done in the other answer), because one first has to define $\frac{\partial f}{\partial\bar{z}}$, and as Ahlfors says, it has no convenient definition as a limit. If one forgets about this, it is easy to fool oneself into thinking that $\frac{\partial f}{\partial z}=0$. But in fact $\frac{\partial f}{\partial z}$ is equal to the complex derivative $f'$ of $f$; this is one reason why this formalism is convenient.
