Question

Evaluate in standard form: $\frac{(2-3i)}{(2-2i)}$, where $i^2=-1$.

• A. $\frac{5}{4}-\frac{i}{4}$
• B. $\frac{5}{4}+\frac{i}{4}$
• C. $-\frac{5}{4}-\frac{i}{4}$
• D. $-\frac{5}{4}+\frac{i}{4}$

Answer 1

The correct option is A $\frac{5}{4}-\frac{i}{4}$

Consider $\frac{(2-3i)}{(2-2i)}$

Rationalise and solve considering that $i^2=-1$ as shown below:

$\Rightarrow \frac{2-3i}{2-2i}\times \frac{2+2i}{2+2i}=\frac{4+4i-6i-{6i}^2}{4-{4i}^2}$

$=\frac{4-2i+6}{4+4}$

$=\frac{-2i+10}{8}$

$=\frac{-i}{4}+\frac{5}{4}$