for two non zero complex polynomial $p(z),q(z)$ we have $p(z)\overline{q(z)}$ is analytic if and only if ?? CSIR - June $2013$ Question is :
for two non zero complex polynomial $p(z),q(z)$ we have $p(z)\overline{q(z)}$ is analytic if and only if 


*

*$p(z)$ is Constant

*$p(z)q(z)$ is Constant

*$q(z)$ is Constant 

*$\overline{p(z)}q(z)$ is Constant


I could easily eliminate first case.


*

*Put $p(z)=1$ and $q(z)=z$ then we would have $p(z)\overline{q(z)}=\bar{z}$ which is not analytic though $p(z)$ is constant.


I could see third case is almost true .


*

*Suppose $q(z)$ is constant then I will be left with only polynomial case so $p(z)\overline{q(z)}$ is analytic. Now, suppose $q(z)$ is not constant then I would have a contradiction which is already considered $q(z)=z$


I could not go any further on second and fourth cases.
Please help me to see this in more detail.
Thank you.
 A: If $p(z)\cdot q(z)$ is constant, then both $p(z)$ and $q(z)$ must be constant, since $\deg (pq) = \deg p + \deg q = 0$, so the second condition is sufficient, but not necessary. 
The third condition - $q$ constant - is sufficient and, since $p \not\equiv 0$, also necessary.
The fourth condition, like the second, implies that both $p$ and $q$ are constant, and hence is also sufficient but not necessary.

Both, $p(z)$ and $q(z)$ are real differentiable and satisfy the Cauchy-Riemann equations,
$$\frac{\partial g}{\partial x} = \frac{\partial h}{\partial y};\qquad \frac{\partial g}{\partial y} = -\frac{\partial h}{\partial x},$$
where $g$ resp. $h$ denote the real resp. imaginary part of a real-differentiable function $f$, and $x,y$ are the real coordinates in $\mathbb{C}\cong \mathbb{R}^2$.
With $q(z)$, also $\overline{q(z)}$ is real-differentiable, and hence also $p(z)\overline{q(z)}$. Writing $p(z) = a(z)+ib(z)$, and $q(z) = c(z) + id(z)$, with real-valued functions $a,b,c,d$, we have
$$p(z)\overline{q(z)} = \bigl(a(z)+ib(z)\bigr)\bigl(c(z)-id(z)\bigr) = \bigl(a(z)c(z) + b(z)d(z)\bigr) + i\bigl(b(z)c(z) - a(z)d(z)\bigr).$$
The Cauchy-Riemann equations in this instance become
$$\frac{\partial(ac +bd)}{\partial x} = \frac{\partial (bc-ad)}{\partial y};\qquad \frac{\partial (ac+bd)}{\partial y} = \frac{\partial (bc-ad)}{\partial x}.$$
By the product rule and linearity,
$$\begin{align}
\frac{\partial(ac+bd)}{\partial x} &= c\frac{\partial a}{\partial x} + a\frac{\partial c}{\partial x} + d\frac{\partial b}{\partial x} + b \frac{\partial d}{\partial x}\\
\frac{\partial (bc-ad)}{\partial y} &= c\frac{\partial b}{\partial y} - a\frac{\partial d}{\partial y} - d\frac{\partial a}{\partial y} + b\frac{\partial c}{\partial y}.
\end{align}$$
Since $p$ and $q$ satisfy the Cauchy-Riemann equations, some terms cancel, and we are left with
$$\frac{\partial (ac+bd)}{\partial x} - \frac{\partial (bc-ad)}{\partial y} = 2a\frac{\partial c}{\partial x} + 2b\frac{\partial d}{\partial x}.$$
Similarly,
$$\begin{align}
\frac{\partial (ac+bd)}{\partial y} &+ \frac{\partial (bc-ad)}{\partial x}\\
&= a\left(\frac{\partial c}{\partial y}-\frac{\partial d}{\partial x}\right)
+ b\left(\frac{\partial d}{\partial y} + \frac{\partial c}{\partial x}\right)
+ c\left(\frac{\partial a}{\partial y} +\frac{\partial b}{\partial x} \right)
+ d\left(\frac{\partial b}{\partial y} - \frac{\partial a}{\partial x}\right)\\
&= 2a \frac{\partial c}{\partial y} + 2b \frac{\partial d}{\partial y}.
\end{align}$$
So the Cauchy-Riemann equations are satisfied for $p(z)\overline{q(z)}$ if and only if
$$\begin{pmatrix}\frac{\partial c}{\partial x} & \frac{\partial d}{\partial x}\\ \frac{\partial c}{\partial y} & \frac{\partial d}{\partial y}\end{pmatrix} \begin{pmatrix} a\\ b\end{pmatrix} \equiv 0.$$
The determinant of the matrix is $\frac{\partial c}{\partial x}\frac{\partial d}{\partial y} - \frac{\partial c}{\partial y}\frac{\partial d}{\partial x} = c_x^2 + c_y^2 = d_x^2 + d_y^2$, which is zero only where all partial derivatives of $q$ vanish. By assumption, $p(z) \not\equiv 0$, so $p$ has only finitely many zeros, and thus for all bit finitely many points, and then by continuity for all points, we must have $c_x = c_y = d_x = d_y = 0$, hence $q(z) = c(z) + id(z)$ must be constant.
This is much more conveniently done with the Wirtinger derivatives,
$$\frac{\partial}{\partial z} = \frac12\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right);\qquad \frac{\partial}{\partial\overline{z}} = \frac12\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right).$$
In the Wirtinger derivatives, the Cauchy-Riemann equations reduce to $\frac{\partial f}{\partial\overline{z}} = 0$, and using the product rule and $\frac{\partial\overline{f}}{\partial\overline{z}} = \overline{\frac{\partial f}{\partial z}}$, we have
$$\frac{\partial}{\partial\overline{z}}\left(p(z)\overline{q(z)}\right) = \frac{\partial p}{\partial\overline{z}}(z)\overline{q(z)} +p(z)\overline{\frac{\partial q}{\partial z}(z)} = p(z)\overline{q'(z)},$$
so for $p(z)\overline{q(z)}$ to be analytic, we need $p\equiv 0$ or $q' \equiv 0$. As $p \equiv 0$ was ruled out, the necessary and sufficient condition is $q' \equiv 0$, or "$q$ is constant".
