Q.Write the complex number $\dfrac{4+3i}{(2+3i)(4-3i)}$ in the form $a+ib$.

Simplify the denominator first, then rationalise by multiplying by its conjugate.

First simplify the denominator $(2+3i)(4-3i)$:

$(2+3i)(4-3i) = 8 - 6i + 12i - 9i^2 = 8 + 6i + 9 = 17+6i,$

using $i^2=-1$ so $-9i^2=9$.

So the expression becomes $\dfrac{4+3i}{17+6i}$. Multiply numerator and denominator by the conjugate $17-6i$:

$\frac{4+3i}{17+6i}\cdot\frac{17-6i}{17-6i} = \frac{(4+3i)(17-6i)}{17^2+6^2}.$

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