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?
Best Answer
$f(z) = u(x,y) + iv(x,y)$
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
$$\frac{\partial{u_1}}{\partial{x}}(x,y) =\frac{\partial{u}}{\partial{x}}(x,-y), \:\:\:\: \frac{\partial{v_1}}{\partial{x}}(x,y) =-\frac{\partial{v}}{\partial{x}}(x,-y) $$
The partial derivatives with respect to $y$ are related by
$$\frac{\partial{u_1}}{\partial{y}}(x,y) =-\frac{\partial{u}}{\partial{y}}(x,y), \:\:\:\: \frac{\partial{v_1}}{\partial{y}}(x,y) =\frac{\partial{v}}{\partial{y}}(x,y) $$
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.