Prove that if \(z\mapsto f(z)\) is analytic, then \(z \mapsto \bar{f(\bar z)}\) is analytic.
Solution: Cauchy-Riemann
It suffices to show that \(g(z) \da \bar{f(\bar z)}\) satisfies CR. Write \(f=u+iv\), then \[ g(x, y) \da a(x, y) + ib(x, y) = u(x, -y) -i v(x, -y) ,\] so we want to show \(a_x = b_y\) and \(a_y = -b_x\). By the chain rule, \[ a_x &= \del_x (x\mapsto u(x, -y)) = u_x \\ a_y &= \del_x (y\mapsto u(x, y))\circ(y\mapsto -y) = -u_y \\ b_x &= \del_x(x\mapsto -v(x, -y)) = -v_x \\ b_y &= \del_x(y \mapsto - v(x, y))\circ(y\mapsto -y) = v_y .\] Now use CR for \(f\) to write \[ a_x &= u_x = v_y = b_y \\ a_y &= -u_y = v_x = -b_x .\]
Solution: Direct definition
Set \(g(z) \da (f(z^*))^* \da \bar{f(\bar z)}\), we can then show \(g'\) exists: \[ \lim_{h\to 0} {g(z+h) - g(z) \over h} &\da \lim_{h\to 0} {f((z+h)^*)^* - f(z^*)^* \over h^{**}} \\ &= \lim_{h\to 0} {\qty{ f(z^* + h^*) - f(z^*) }^* \over h^{**}} \\ &= \lim_{h\to 0} \qty{ f(z^* + h^* ) - f(z^*) \over h^* }^* \\ &\da \qty{f'(z^*)}^* ,\] where we’ve used that \(w\mapsto w^*\) is continuous to commute a limit. So this limit exists, \(g\) is differentiable with \(g'(z) \da \bar{f'(\bar z)}\).
Solution: Power series
Since \(f\) is analytic, take a Laurent expansion \(f(z) = \sum_{k\geq 0} c_k z^k\). Then \[ g(z) \da (f(z^*))^* = \qty{\sum_{k\geq 0} c_k \bar{z^k} }^* = \sum_{k\geq 0} \bar{c_k} z^k ,\] making \(g\) analytic.