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$\text {Prove that f is analytic if and only if}$ $\frac{\partial f}{\partial \overline{Z}}=0$

MY attempt:

suppose f is analytic $f(z)=u+iv \text{ then } u_x=v_y \text{ and } u_y=-v_x\\ \text{also } x=\frac{z+\overline{z}}{2},y=\frac{z-\overline{z}}{2i}\\ \frac{\partial f}{\partial z}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \overline{z}}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \overline{z}}=0$

but to prove converse part of this statement

Inverse Problem
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    One has $\frac{\partial f}{\partial \bar z}(z_0) =\frac 12\left(\frac{\partial f}{\partial x}(z_0) +i\frac{\partial f}{\partial y}(z_0) \right) $, and the vanishing of this at $z_0$ amounts to the Cauchy--Riemann equations. – Pedro Jul 31 '17 at 03:41
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    See this, too. – Pedro Jul 31 '17 at 03:43

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