Question
Express the following complex numbers in the standard form a + ib: $\frac{(2+\text{i})^3}{2+3\text{i}}$

Answer

$\frac{(2+\text{i})^3}{2+3\text{i}}=\frac{2^3+\text{i}^3+3\times2\times\text{i}(2+\text{i})}{2+3\text{i}} \ \big[\therefore \ (\text{a}+\text{b})^3=\text{a}^3+\text{b}^3+3\text{ab}(\text{a}+\text{b})\big]$ $=\frac{(8-\text{i}+6\text{i}(2+\text{i}))}{2+3\text{i}}\times\frac{(2-3\text{i})}{2-3\text{i}}$ (On rationalising the denominator) $=\frac{(8-\text{i}+12\text{i}+6\text{i}^2)(2-3\text{i})}{2^2+3^3}$ $=\frac{(8-6+11\text{i})(2-3\text{i})}{4+9} \ \big(\because \ \text{i}^2=-1\big)$ $=\frac{(2+11\text{i})(2-3\text{i})}{13}$ $=\frac{4-6\text{i}+22\text{i}+33}{13}$ $=\frac{37+16\text{i}}{13}$ $=\frac{37}{13}+\frac{16}{13}\text{i}$ $\therefore\frac{(2+\text{i})^3}{2+3\text{i}}=\frac{37}{13}+\frac{16}{13}\text{i}$

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