# Given $f(z)$ is an analytic complex function, show $f^*(z^*)$ is also analytic

This exercise is from Arfken mathematical methods for physicists): "The function $f(z)=u(x,y)+iv(x,y)$ is analytic. Show that $f^*(z^*)$ is also analytic."

There must be some simple proof (and not related to series), because there is little said about complex analysis in the book before this exercise (The only important thing said is Cauchy-Riemann conditions). Not sure, but I think if $f(z)=u(x,y)+iv(x,y)$ then $f^*(z^*)=u(x,-y)-iv(x,-y)=g(x,y)+ih(x,y)$. Now $g$ and $h$ must satisfy the Cauchy-Reimann conditions and their first partial derivatives with respect to $x$ and $y$ must be continuous. Yet I can't show any of them. Any suggestions for developing this idea?

• @mrf This is not the same. He wants to show it from the Cauchy-Riemann equations and the question you are saying is a duplicate explicitly say in the answer "I think it will be easiest not to split into real and imaginary parts -- so no Cauchy-Riemann " – JKnecht Mar 26 '16 at 20:02
• It's been asked a dozen times or so. Here's another duplicate: math.stackexchange.com/questions/769099 (but I guess I wasn't able to reopen and close again.) And another: math.stackexchange.com/questions/809145 – mrf Mar 26 '16 at 20:26
• @mrf Got it. Did not know that. – JKnecht Mar 26 '16 at 20:31
• @mrf I asked a new question because none of those questions say how real and imaginary parts of f*(z*) satisfy the Cauchy-Riemann conditions which is my main problem. – Simorq Mar 27 '16 at 7:46

$f(z) = u(x,y) + iv(x,y)$

You want to show that the function

$f^*(z^*) = u(x,-y) - iv(x,-y)$

is also analytic.

Denote $f^*(z^*)$ by

$f^*(z^*) = u_1 + iv_1$

with

$u_1(x,y) = u(x,-y)$

$v_1(x,y) = -v(x,-y)$

The partial derivatives with respect to $x$ are related by

$$\frac{\partial{u_1}}{\partial{x}} =\frac{\partial{u}}{\partial{x}}, \:\:\:\: \frac{\partial{v_1}}{\partial{x}} =-\frac{\partial{v}}{\partial{x}}$$

The partial derivatives with respect to $y$ are related by

$$\frac{\partial{u_1}}{\partial{y}} =-\frac{\partial{u}}{\partial{y}}, \:\:\:\: \frac{\partial{v_1}}{\partial{y}} =\frac{\partial{v}}{\partial{y}}$$

You know that the Cauchy Riemann equations holds for $f(z) = u + iv$. The above relations shows that they also hold for $f^*(z^*) = u_1 + iv_1$ and thus the function $f^*(z^*)$ is also analytic.

• Thanks, but I don't understand any of these. For example why \frac{\partial{u_1}}{\partial{x}} =\frac{\partial{u}}{\partial{x}} – Simorq Mar 26 '16 at 19:15
• @Simorq I have edited my answer. Do you understand now? – JKnecht Mar 26 '16 at 19:57
• @Simorq You just plug the above expressions for the partial derivatives into the Cauchy Riemann equations. – JKnecht Mar 26 '16 at 20:05
• How did you find those relations between partial derivatives of u and u1? This is what I don't understand. – Simorq Mar 27 '16 at 7:50
• @Simorq I used these $$u_1(x,y) = u(x,-y), \:\: \:\: v_1(x,y) = -v(x,-y)$$ – JKnecht Mar 27 '16 at 8:28