Show $f(z)=2u\left( \frac z2, \frac{-iz}{2}\right ) + \text{constant} $ if $f(z)=u(x,y)+iv(x,y)$ is an analytic function Suppose  $f(z)=u(x,y)+iv(x,y)$  be an analytic function. Show that
$ \displaystyle (a)\;  f(z)=2u\left(\frac z2,\frac{-iz}{2} \right ) +\text{ constant} \\(b) \; f(z)=2iv\left(\frac z2,\frac{iz}{2} \right )+\text{ constant}$ 
 A: Perhaps you need some hypotheses?  For example
$$
|z| = \sqrt{x^2+y^2} + i\cdot 0;\quad f(z):=|z|, u(x,y) := \sqrt{x^2+y^2}, v(x,y):=0
$$
but
$$
2u(z/2,-iz/2) = 2\sqrt{\left(\frac{z}{2}\right)^2+\left(\frac{-iz}{2}\right)^2}
=2\sqrt{\frac{z^2-z^2}{4}}=0
$$
A: Part (a) of your question is discussed in Ahlfors' "Complex Analysis" text on page 27. He warns ahead of the discussion that it is purely formal and shouldn't be used to prove anything.
First of all, you need to consider the formal derivatives:
$$\frac{\partial f}{\partial z}=\frac{1}{2} \left(\frac{\partial f}{\partial x}-i \frac{\partial f}{\partial y} \right), \frac{\partial f}{\partial \bar{z}}=\frac{1}{2} \left(\frac{\partial f}{\partial x}+i \frac{\partial f}{\partial y} \right). $$
Note that the Cauchy-Riemann equations are simply $\frac{\partial f}{\partial \bar{z}}=0$.
Take any analytic function $f(z)=u(x,y)+iv(x,y)$, Its conjugate $\overline{f(z)}$ satisfies $\frac{\partial \overline{f(z)}}{\partial z}=0$. Formally this means that $\overline{f(z)}$ is a function of $\bar{z}$ only; Denote this function by $\bar{f}(\bar{z})$. Thus we have 
$$u(x,y)=\frac{1}{2} \left( f(x+iy)+\bar{f}(x-iy) \right).$$
Once again, we formally interpret this identity for complex $x,y$. We can therefore plug in $x=\frac{z}{2},y=\frac{z}{2i}$, and obtain:
$$u \left(\frac{z}{2},\frac{z}{2i} \right)=\frac{1}{2} \left( f(z)+\bar{f}(0) \right), $$ which is upon rearrangement $$f(z)=2u \left(\frac{z}{2},\frac{z}{2i} \right)+\text{const}. $$
Part (b) can be done similarly.
The greatest issue here is plugging in complex values for $x,y$. In general it cannot be done, but for simple functions, such as polynomials, everything is OK.
