Question

Simlify: $\sqrt{\frac{-17}{144}-i}$

• A. $\pm \left(\frac{3}{2}-\frac{i}{3}\right)$
• B. $\pm \left(\frac{3}{4}-\frac{2i}{3}\right)$
• C. $\pm \left(\frac{3}{5}-\frac{5i}{6}\right)$
• D. $\pm \left(\frac{2}{3}-\frac{3i}{4}\right)$

Answer 1

The correct option is D $\pm \left(\frac{2}{3}-\frac{3i}{4}\right)$

We know that,

${(a-b)}^2=a^2+b^2-2ab$

Now, let,

$-2ab=-i$

$ab=i$

Now, consider $\frac{-17}{144}$. We can write it as,

$\frac{-17}{144}=\frac{64-81}{9\times 16}=\frac{4}{9}-\frac{9}{16}$

Thus,

$\sqrt{\frac{-17}{144}-i}=\sqrt{\frac{4}{9}-\frac{9}{16}-i}$

$=\sqrt{{\left(\frac{2}{3}\right)}^2+{\left(i\right)}^2{\left(\frac{3}{4}\right)}^2-2\times \left(\frac{2}{3}\right)\times \left(\frac{3i}{4}\right)}$

$=\sqrt{{(\frac{2}{3}-\frac{3i}{4})}^2}$

$=\pm (\frac{2}{3}-\frac{3i}{4})$

Hence, this is the required result.