Question
Express the following complex numbers in the standard form a + ib: $\frac{2+3\text{i}}{4+5\text{i}}$

Answer

$\frac{2+3\text{i}}{4+5\text{i}}=\frac{2+3\text{i}}{4+5\text{i}}\times\frac{(4-5\text{i})}{(4-5\text{i})}$ (On rationalising the denominator)
$=\frac{2(4-5\text{i})+3\text{i}(4-5\text{i})}{4^2+5^2}$
$=\frac{8-10\text{i}+12\text{i}+15}{16+25} \ \big(\because\ \text{i}^2=-1\big)$
$=\frac{23+2\text{i}}{41}$
$=\frac{23}{41}+\frac{2}{41}\text{i}$
$\therefore \ \frac{2+3\text{i}}{4+5\text{i}}=\frac{23}{41}+\frac{2}{41}\text{i}$

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