<s>[INST] Given that $T=\frac{x-iy}{x+iy}$, where $x, y, T \in\Bbb R$, show that $\frac{1+T^2}{2T}=\frac{x^2-y^2}{x^2+y^2}$ Given that $T=\frac{x-iy}{x+iy}$, where $x, y, T \in\Bbb R$, show that $$\frac{1+T^2}{2T}=\frac{x^2-y^2}{x^2+y^2}$$
I have got rid of $i$ in the denominator, leaving $$\frac{x^2-y^2-2ixy}{x^2+y^2}$$
but now I am stuck. What are the next steps please?
 [/INST] It may be easier to note that our expression is equal to 
$$\frac{1}{2}\left(\frac{1}{T}+T\right).$$
Substituting, we get
$$\frac{1}{2}\left(\frac{x+iy}{x-iy}+\frac{x-iy}{x+iy}    \right).$$
Bring to the common denominator $x^2+y^2$. The top is then $(x+iy)^2+(x-iy)^2=2x^2-2y^2$, and we are finished.
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