If real parts of $\sqrt{-5-12 i}, \sqrt{5+12 i}$ are positive values, the real part of $\sqrt{-8-6 i}$ is a…
If real parts of $\sqrt{-5-12 i}, \sqrt{5+12 i}$ are positive values, the real part of $\sqrt{-8-6 i}$ is a negative value and $a+i b=\frac{\sqrt{-5-12 i}+\sqrt{5+12 i}}{\sqrt{-8-6 i}}$ then $2 a+b=$
$3$
$2$
$-3$
$-2$
Solution
Let $\sqrt{-5-12 i}=a+b i, a\gt0$
$\Rightarrow-5-12 i=(a+b i)^2$
So $a^2-b^2=-5$ and $2 a b=-12$
Now, $a^2+b^2=\sqrt{\left(a^2-b^2\right)^2+4 a^2 b^2}$ $=\sqrt{(-5)^2+(-12)^2}=\sqrt{169}=13$
$\begin{aligned} & \left\{a^2-b^2=-5, a^2+b^2=13\right\} \Rightarrow 2 a^2=8 \Rightarrow a=2 \quad(\because a\gt0) \\ & \text { and } b=\frac{-12}{4}=-3 . \text {So, } \sqrt{-5-12 i}=2-3 i\end{aligned}$
Similarly, we get $\sqrt{5+12 i}=3+2 i$ and $\sqrt{-8-6 i}=-1+3 i$
Now, $a+b i=\frac{\sqrt{-5-12 i}+\sqrt{5+12 i}}{\sqrt{-8-6 i}}$
$\begin{aligned} & \text {Now, } a+b i=\frac{\sqrt{-5-12 i}+\sqrt{5+12 i}}{\sqrt{-8-6 i}} \\ & =\frac{2-3 i+3+2 i}{-1+3 i}=\frac{5-i}{-1+3 i} \times \frac{-1-3 i}{-1-3 i}\end{aligned}$
$=\frac{-8-14 i}{10}=\frac{-4}{5}-\frac{7}{5} i \Rightarrow a=\frac{-4}{5} \text { and } b=\frac{-7}{5}$
So, $2 a+b=\frac{-8}{5}-\frac{7}{5}=\frac{-15}{5}=-3$