I am reading about complex analysis on my own. I came across this result which is equivalent to Cauchy Riemann equations.
A necessary condition for a complex valued function of complex variable to be differentiable at a point is
$$\frac{df}{d\bar{z}}=0$$
I am trying to practice through the examples.
Please have a look and point out the error.
Note: I will denote $\frac{df}{d\bar{z}}=f'$
Problem 1 $f(z)=\bar{z}$
Solution: $f'=1$ so nowhere differentiable.
Problem 2 $f(z)=\Re(z)$
$2 \Re(z)=z+\bar{z}$, so $f'=1/2$. We conclude that function is nowhere differentiable.
Problem 3, $f(z)=|z|$
$f= \sqrt{z\bar{z}}$
$f'=\frac{z}{2\sqrt{z\bar{z}}}$. We conclude that function is not differentiable for $z\neq0$.
Problem 4 $f(z)=|z|^{2}$
As $|z|^{2}=z\bar{z}$, we get $f'=z$, so function is not differentiable at $z\neq 0$
Problem 5 $f(z)=\Re(z)^2$
$f=\frac{z+\bar{z}}{2}^2=\frac{1}{4}(z^2+\bar{z}^2+2z\bar{z})$
So $f'=\frac{z+\bar{z}}{2}=\Re(z)$
Now this implies that function is not differentiable at all points $z$ where $\Re(z)\neq0$
Related Questions
Since it is just a necessary condition, we can not conclude whether a function is differentiable at the points where $f'=0$. We have to use the definition of differentiability.
I know, form the theory of Cauchy Riemann Equations that if partial derivative exist and is continuous then CR equations become a sufficient condition. How to capture this point in $\frac{df}{d\bar{z}}=0$
Best Answer
With the definition $$ {\partial\over \partial \overline{z}}=\frac12\left({\partial\over \partial x}+i{\partial\over \partial y}\right), $$ the equation $$ {\partial f\over \partial \overline{z}} = 0 \tag{1}$$ is just a restatement of the Cauchy-Riemann equations, so that it has the same requirements for sufficiency. According to Wikipedia it is enough that the real and imaginary parts of $f$ be real-differentiable, so you don't have to assume continuity of the partial derivatives.
I'm not sure I understand the very last sentence of you question, but I would tentatively say that there is no way to capture the sufficient condition in $(1)$ any more than it's possible to capture it in the C-R equations, because $(1)$ is identical to the C-R equations.