Question:
Express $\frac{3-\sqrt{-16}}{1-\sqrt{-9}}$ in the form $(a+i b)$.
Solution:
We have, $\frac{3-\sqrt{-16}}{1-\sqrt{-9}}$
We know that $\sqrt{-1}=\mathrm{i}$
Therefore,
$\frac{3-\sqrt{-16}}{1-\sqrt{-9}}=\frac{3-4 i}{1-3 i}$
$\frac{3-\sqrt{-16}}{1-\sqrt{-9}}=\frac{3-4 i}{1-3 i} \times \frac{1+3 i}{1+3 i}$
$\frac{3-\sqrt{-16}}{1-\sqrt{-9}}=\frac{3+9 i-4 i-12 i^{2}}{(1)^{2}-(3 i)^{2}}$
$\frac{3-\sqrt{-16}}{1-\sqrt{-9}}=\frac{15+5 i}{1+9}=\frac{15}{10}+\frac{5 i}{10}=\frac{3}{2}+\frac{1}{2} i$
Hence,
$\frac{3-\sqrt{-16}}{1-\sqrt{-9}}=\frac{3}{2}+\frac{i}{2}$
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