Question

If $a>0,|z|=$ a then find the real part of $\left(\frac{2-a}{z+a}\right).$

Answer 1

Let $z=x+iy⟹a=\sqrt{x^2+y^2}$

$⟹\frac{2-a}{x+iy+a}$

Multiplying and dividing by $x+a-iy$

$⟹\frac{2-a}{x+iy+a}\times \frac{x+a-iy}{x+a-iy}$

$⟹\frac{(2-a)(x+a)-i(2-a)y}{(x+a{)}^2+y^2}$

So, Real part of fraction is $\frac{(2-a)(x+a)}{(x+a{)}^2+y^2}$

$⟹\frac{2x+(2+x)a-a^2}{x^2+a^2+2ax+y}$