How should I structure this complex analysis proof? 
Suppose that $f$ is an entire function (i.e. analytic over the entire complex plane). Show that if
  $$f(z)=\overline{i f(-z)}$$
  then $f$ must be a constant function of the form
  $$f(z)=c(1-i)$$
  for some real number $c .$ (HINT: Begin by showing that $f$ must be a constant. Once you know that, the form of the constant is easy to prove. To show that $f$ is a constant function, write $f=u+i v$ as usual and compare the results of the Cauchy-Riemann equations for $u$ and $v$ with the equations you get from the relation above. You should be trying to show that both partial derivatives bf $u$ and $v$ are zero.

I began by rewriting $f(z)=u+iv$ and $if(-z)$ as $conjugate(-v+iu) = -v-iu$
The left-hand side has the Cauchy-Riemann equations $u_x=\frac{\partial u}{\partial x}, u_y=\frac{\partial u}{\partial y}$ and the right-hand side $u_x=-i\frac{\partial u}{\partial x}, u_y=-i\frac{\partial u}{\partial y}$.
I noticed $\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial x} = \frac{\partial u}{\partial x}(1-i)$ and $-\frac{\partial u}{\partial y}-i\frac{\partial u}{\partial y}=-\frac{\partial u}{\partial y}(1-i)$.
I know I haven't completed the necessary steps to justify the conclusion. How can this information help me? Disclaimer: I know the CR equations are $u_x=v_y$ and $u_y=-v_x$, I just didn't type them fully above. Any help is greatly appreciated.
 A: Say, $f(z)=u(z) +iv(z)$ then, $\overline{if(-z)}=-iu(-z)-v(-z)$ and and so using the relation $g'(z)=\frac{1}{2}(\frac{\partial g}{\partial x}(z)-i\frac{\partial g}{\partial y}(z))$ for holomorphic functions  we have that  $\overline{if(-z)}'= -i\partial_xu(-z)  -\partial_xv(-z) -\partial_yu(-z) +i\partial_yv(-z)$.Thus using the CR equations for $f(z)$ we get that $f'(z)=\overline{if(-z)}'=0$.Thus $f$ is constant and the particular form can be easily determined now.
A: You know that $u=-v$. So, the Cauchy-Riemann equation $u_x=v_y$ becomes $u_x=-u_y$. On the other hand, $u_y=-v_x=u_x$. But then, since $u_x$ is equal to both $u_y$ and to $-u_y$, $u_x$ and $u_y$ are both the null function. So, by the Cauchy-Riemann equations, $v_x$ are $v_y$ are the null function too. So, $u$, and $v$ are constant, and so $f$ is equal to a constant $C=x+yi$ ($x,y\in\Bbb R$). Since $C=\overline{iC}$, $x+yi=-y-xi$. So, $x=-y$ and therefore $C=c(1-i)$ for some $c\in\Bbb R$.
A: We have $f=u+vi=-v-ui$. By just matching the real and imaginary parts of both sides, we have
\begin{align}
u_x=-v_x \\
u_y=-v_y 
\end{align}
By CR, you also know $u_x=v_y \implies v_y=-v_x$ and $u_y=-v_x\implies -v_x=-v_y\implies v_x=v_y.$ Thus, $v_y=v_x$ and $v_y=-v_x$, which is only possible for $v_y=v_x=0$ so $v\equiv c_v$.
But then from CR, you also know $u_x=u_y=0$ so $u\equiv c_u$. Now we know $f=c_u+c_v i$. From here, go back to the original equation in that 
\begin{align}
c_u+c_v i=-c_v-c_ui
\end{align}
Clearly, $c_u=-c_v$ so $f=c_u-c_u i = c_u(1-i)$.
