conjugate complex numbers (ahlfors) I am solving the ahlfors for my course. I have some questions:
1.Verify the calculation that the values of $\dfrac{z}{z^{2}+1}$ for $z=x+iy$ and $z=x-iy$ are conjugate.
The first step I calculated what the value is $\dfrac{2}{z^{2}+1}$ for $z=x+iy$
Then I did the same thing for $z=x+iy$
But the solutions are different. Might I do something wrong?


*The question is: "what is the solution of $z^{2}+\left( \alpha +i{\beta }\right) z+\gamma+i\delta =0$" 


First I tried to solve this equation for putting $z=x+iy$. But in so many numbers I couldn't get what to find. I think I need to find x and y but I got lost.

For $z$, we have that $z^2 = x^2 - y^2 + 2xyi$.
  \begin{align*}                                                                
    \frac{z}{z^2 + 1}                                                           
    & = \frac{x + iy}{x^2 - y^2 + 1 + 2xyi}\\                                   
    & = \frac{x + iy}{x^2 - y^2 + 1 + 2xyi}                                     
      \frac{x^2 - y^2 + 1 - 2xyi}{x^2 - y^2 + 1 - 2xyi}\\                       
    & = \frac{x(x^2 - y^2 + 1) + 2xy^2 + iy(x^2 - y^2 + 1 - 2x^2)}              
      {(x^2 - y^2 + 1)^2 + 4x^2y^2}\tag{1}                    
  \end{align*}
  For $\bar{z}$, we have that $\bar{z}^2 = x^2 - y^2 - 2xyi$.
\begin{align*}                                                                
    \frac{\bar{z}}{\bar{z}^2 + 1}                                               
    & = \frac{x - iy}{x^2 - y^2 + 1 - 2xyi}\\                                   
    & = \frac{x - iy}{x^2 - y^2 + 1 - 2xyi}                                     
      \frac{x^2 - y^2 + 1 + 2xyi}{x^2 - y^2 + 1 + 2xyi}\\                       
    & = \frac{x(x^2 - y^2 + 1) + 2xy^2 - iy(x^2 - y^2 + 1 - 2x^2)}              
      {(x^2 - y^2 + 1)^2 + 4x^2y^2}\tag{2}                 
  \end{align*}
  Therefore, we have that equations $(1)$ and $(2)$ are conjugates.

I asked this problem here on Dec 18, 2014 wondering if it could be simplified more but I was told it couldn't be reduced in further.
The quadratic equation is $x = \frac{-b\pm\sqrt{b^2 - ac}}{2}$.
  For the complex polynomial, we have
  $$                                                                            
  z = \frac{-\alpha - \beta i\pm                                                
    \sqrt{\alpha^2 - \beta^2 - 4\gamma + i(2\alpha\beta - 4\delta)}}{2}         
  $$
  Let
  $a + bi = \sqrt{\alpha^2 - \beta^2 - 4\gamma + i(2\alpha\beta - 4\delta)}$.
  Then
  $$                                                                          
  z = \frac{-\alpha - \beta\pm (a + bi)}{2}                                     
  $$
