Question

Express $\left(\frac{2+3i}{2-3i}\right)-\left(\frac{2-3i}{2+3i}\right)$ in the term of $a+ib$

• A. $\frac{24i}{13}$
• B. $\frac{27i}{13}$
• C. $\frac{28i}{13}$
• D. $\frac{30i}{13}$

Answer 1

The correct option is A $\frac{24i}{13}$

We have,

$\begin{array}{c}\left(\frac{2+3i}{2-3i}\right)-\left(\frac{2-3i}{2+3i}\right)\\ \left(\frac{2+3i}{2-3i}\right)\left(\frac{2+3i}{2+3i}\right)-\left(\frac{2-3i}{2+3i}\right)\left(\frac{2-3i}{2-3i}\right)\\ \frac{{\left(2+3i\right)}^2}{4+9}-\frac{{\left(2-3i\right)}^2}{4+9}\\ \Rightarrow \frac{4-9+12i-\left(4-9-12i\right)}{13}\\ =\frac{-4-9+12i-4+9+12i}{13}=\frac{24i}{13}\\ a+;b=0+\frac{24i}{13}\end{array}$

Hence, this is the answer.