Question:
Express each of the following in the form (a + ib):
$\frac{1}{(4+3 \mathrm{i})}$
Solution:
Given: $\frac{1}{4+3 i}$
Now, rationalizing
$=\frac{1}{4+3 i} \times \frac{4-3 i}{4-3 i}$
$=\frac{4-3 i}{(4+3 i)(4-3 i)} \ldots(\mathrm{i})$
Now, we know that,
$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
So, eq. (i) become
$=\frac{4-3 i}{(4)^{2}-(3 i)^{2}}$
$=\frac{4-3 i}{16-9 i^{2}}$
$=\frac{4-3 i}{16-9(-1)}\left[\because i^{2}=-1\right]$
$=\frac{4-3 i}{16+9}$
$=\frac{4-3 i}{25}$
$=\frac{4}{25}-\frac{3}{25} i$
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