Question

$\frac{\sqrt{6+8i}+\sqrt{6-8i}}{\sqrt{6+8i}-\sqrt{6-8i}}$ is equal to ( where $i=\sqrt{-1}$

• A. $2i$
• B. $i$
• C. $-2i$
• D. $-i$

Answer 1

The correct option is C $-2i$

$\frac{\sqrt{4+8i}+\sqrt{6-8i}}{\sqrt{6+8i}-\sqrt{6-8i}}$

On rationalising we get

$\Rightarrow \frac{(\sqrt{6+8i}+\sqrt{6-8i})(\sqrt{6+8i}+\sqrt{6-8i})}{(\sqrt{6+8i}{)}^2-(\sqrt{6-8i}{)}^2}$

$\Rightarrow \frac{(\sqrt{6+8i}{)}^2+(\sqrt{6-8i}{)}^2+2\left(\sqrt{6+8i}\right)\left(\sqrt{6-8i}\right)}{6+8i-6+8i}$

$\Rightarrow \frac{6+8i+6-8i+2\sqrt{36-64i^2}}{16i}$

$=\frac{12+2\times 10}{16i}$

$=\frac{232}{18i}=\frac{-2\times i^2}{i}=-2i$.