# $\overline{f(\overline{z})}$ and Cauchy-Riemann Equations

Given that $f(z)=u(x,y)+iv(x,y)$ and Cauchy-Riemann equations hold for $f$ at a point $z_0=x_0+iy_0$ what can be said about $g(z)=\overline{f(\overline{z})}=u(x,-y)-iv(x,-y)$ at the same point?

I tried to solve that using the chain rule defining $w=-y$ but it leads to a wrong answer.

$$\frac{\partial{g}}{\partial{\overline{z}}}=\frac{1}{2}\bigg(\frac{\partial{g}}{\partial{x}}+i\frac{\partial{g}}{\partial{y}}\bigg)=0$$

$$\frac{\partial{g}}{\partial{x}}=\frac{\partial{g}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{g}}{\partial{v}}\frac{\partial{v}}{\partial{x}}=\frac{\partial{u}}{\partial{x}}-i\frac{\partial{v}}{\partial{x}}$$

$$\frac{\partial{g}}{\partial{y}}=\frac{\partial{g}}{\partial{u}}\frac{\partial{u}}{\partial{w}}\frac{\partial{w}}{\partial{y}}+\frac{\partial{g}}{\partial{v}}\frac{\partial{v}}{\partial{w}}\frac{\partial{w}}{\partial{y}}=-\frac{\partial{u}}{\partial{w}}+i\frac{\partial{v}}{\partial{w}}$$

$$2\frac{\partial{g}}{\partial{\overline{z}}}=\bigg(\frac{\partial{u}}{\partial{x}}-\frac{\partial{v}}{\partial{w}}\bigg)-i\bigg(\frac{\partial{v}}{\partial{x}}+\frac{\partial{u}}{\partial{w}}\bigg)=0$$

Since Cauchy-Riemann hold for $f$, $\frac{\partial{u}}{\partial{x}}=\frac{\partial{v}}{\partial{y}}$ and $\frac{\partial{v}}{\partial{x}}=-\frac{\partial{u}}{\partial{y}}$:

$$\frac{\partial{v}}{\partial{y}}=\frac{\partial{v}}{\partial{w}}$$

$$\frac{\partial{u}}{\partial{y}}=\frac{\partial{u}}{\partial{w}}$$

That implies Cauchy Riemann equations don't hold for $g$ at $z_0$ since $w=-y$ which is wrong. I would like to know why the above approach is failing.

• this statement is not correct $$\frac{\partial{g}}{\partial{x}}=\frac{\partial{g}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{g}}{\partial{v}}\frac{\partial{v}}{\partial{x}}=\frac{\partial{u}}{\partial{x}}-i\frac{\partial{v}}{\partial{x}}$$ – Guy Fsone Oct 7 '17 at 15:51
• I didn't get it. I thought it was right because: g -> u -> x and g -> v -> x (chain rule "tree" diagram) – Zalnd Oct 7 '17 at 16:34
• u is a function not a variable for $g$ – Guy Fsone Oct 7 '17 at 16:59
• I think dg/dx = d/dx * ( u(x,-y) - i*v(x,-y) ) also leads to the same equation. – Zalnd Oct 7 '17 at 19:49