Question

If $\frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}=a+ib,b<0$, then the value of $a-2b$ is

Answer 1

Given : $\frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}=a+ib$
$\frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}=\frac{{(\sqrt{5+12i}+\sqrt{5-12i})}^2}{(5+12i)-(5-12i)}=\frac{5+12i+5-12i+2\sqrt{(5+12i)}\sqrt{(5-12i)}}{24i}$

We know that,
$\sqrt{a+ib}=\pm \left[\sqrt{\frac{1}{2}(\sqrt{a^2+b^2}+a}\right)+i\sqrt{\frac{1}{2}(\sqrt{a^2+b^2}-a}\left)\right]\therefore \sqrt{5+12i}=\pm (3+2i)$
Similarly,
$\sqrt{5-12i}=\pm (3-2i)\therefore a+ib=\frac{10\pm 2(3+2i)(3-2i)}{24i}\Rightarrow a+ib=\frac{10\pm 26}{24i}\Rightarrow a+ib=\frac{-3i}{2}\text{and}a+ib=\frac{2i}{3}$
But as given in question $b<0$
Hence $a=0,b=\frac{-3}{2}$
So, value of $a-2b=3$