Suppose $f = u + iv \in H(\mathbb{C})$ satisfies $u(x, y) = u(−y, x)$. Show that $f (z) = f (iz)$ for all $z \in \mathbb{C}$ 
Suppose $f = u + iv \in H(\mathbb{C})$ satisfies $u(x, y) = u(−y, x)$. Show that $f (z) = f (iz)$ for all $z \in \mathbb{C}$.

How to solve this ? I tried so many times but I could not find any way to prove this .
 A: Hint: The given condition shows that the real part of the entire function $f(iz)-f(z)$ vanishes identically.
A: Here to show that $f(z)=f(iz)$ we need to show that $v(x,y)=v(-y,x)$ as $u(x,y)=u(-y,x)$ is already given. Now due to cauchy-riemann equations. $u_x(x,y) = v_y(x,y)$ and $u_y(x,y) = -v_x(x,y) $. But, $u_x(x,y)=u_x(-y,x)$ and therefore $v_y(x,y) = u_x(-y,x)$. Now, by Cauchy Riemann equations for $f(iz)$, we get that $u_x(-y,x)=v_y(-y,x)$. Combining with the last equality we get $$v_y(x,y) = v_y(-y,x) $$ By proceeding on exactly similar lines with the other equality we get $$v_x(x,y)=v_x(-y,x)$$. Both these equalities show that both the partial derivatives of $v(x,y)$ and $v(-y,x)$ are equal everywhere. Try to show by integrating and differentiating once that $v(x,y)=v(-y,x)$.(Since you are dealing with partial derivatives you will have to add $g(y)$ if integrating wrt $x$ and vice-versa.)
A: Suppose that $f(z)=f(x+iy)=u(x,y)+iv(x,y)$ is analytic.  Then, $u$ and $v$ satisfy the Cauchy-Riemann equations
$$\begin{align}
\frac{\partial u(x,y)}{\partial x}&=\frac{\partial v(x,y)}{\partial y} \tag 1\\\\
\frac{\partial u(x,y)}{\partial y}&=-\frac{\partial v(x,y)}{\partial x}\tag2
\end{align}$$

Next, suppose that $u(x,y)=u(-y,x)$.  Then, using $(1)$ followed by using $(2)$ reveals
$$\begin{align}
\frac{\partial v(x,y)}{\partial y}&=\frac{\partial u(x,y)}{\partial x}\\\\
&=\frac{\partial u(-y,x)}{\partial x}\\\\
&= -\frac{\partial v(-y,x)}{\partial (-y)}\\\\
&=\frac{\partial v(-y,x)}{\partial y}\tag 3
\end{align}$$ 
From $(3)$ we see that $v(x,y)=v(-y,x)+\phi(x)$ for some function $\phi(x)$.

Similarly, using $(2)$ followed by using $(1)$ reveals
$$\begin{align}
\frac{\partial v(x,y)}{\partial x}&=-\frac{\partial u(x,y)}{\partial y}\\\\
&=-\frac{\partial u(-y,x)}{\partial y}\\\\
&=\frac{\partial u(-y,x)}{\partial (-y)}\\\\
&= \frac{\partial v(-y,x)}{\partial x}\\\\
&\tag 4
\end{align}$$ 
From $(4)$ we find that $v(x,y)=v(-y,x)+\psi(y)$ for some function $\psi(y)$.

Since $\phi(x)=\psi(y)$, then both functions must be independent of $x$ and $y$ (i.e., a constant, say $C$). Hence, $v(x,y)=v(-y,x)+C$. At $(x,y)=(0,0)$, $v(0,0)=v(0,0)+C$ implies $C=0$ and thus $v(x,y)=v(-y,x)$.
Finally, we find that 
$$f(z)=u(x,y)+iv(x,y)=u(-y,x)+iv(-y,x)=f(iz)$$
as was to be shown!
