Answer
$\dfrac{1}{26}+\dfrac{3}{52}i$
Work Step by Step
The reciprocal of the given complex number, $
8-12i
,$ is
\begin{align*}\require{cancel}
&
\dfrac{1}{8-12i}
.\end{align*}
Multiplying the numerator and the denominator by the conjugate of the denominator, the expression above is equivalent to
\begin{align*}
&
\dfrac{1}{8-12i}\cdot\dfrac{8+12i}{8+12i}
\\\\&=
\dfrac{8+12i}{(8-12i)(8+12i)}
.\end{align*}
Using $(a+b)(a-b)=a^2-b^2$ or the special product of the sum and difference of like terms, the expression above is equivalent to
\begin{align*}
&
\dfrac{8+12i}{(8)^2-(12i)^2}
\\\\&=
\dfrac{8+12i}{64-144i^2}
\\\\&=
\dfrac{8+12i}{64-144(-1)}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{8+12i}{64+144}
\\\\&=
\dfrac{8+12i}{208}
\\\\&=
\dfrac{\cancel8^2+\cancel{12}^3i}{\cancel{208}^{52}}
&\text{ (divide by $4$)}
\\\\&=
\dfrac{2+3i}{52}
.\end{align*}
The reciprocal is $
\dfrac{2+3i}{52}
$.
Checking:
Multiplying the given expression, $
8-12i
$ and the reciprocal, $
\dfrac{2+3i}{52}
,$ results to
\begin{align*}
&
(8-12i)\cdot\left( \dfrac{2+3i}{52} \right)
\\\\&=
\dfrac{(8-12i)(2+3i)}{52}
\\\\&=
\dfrac{4(2-3i)(2+3i)}{52}
\\\\&=
\dfrac{\cancel4^1(2-3i)(2+3i)}{\cancel{52}^{13}}
&\text{ (divide by $4$)}
\\\\&=
\dfrac{(2-3i)(2+3i)}{13}
\\\\&=
\dfrac{(2)^2-(3i)^2}{13}
&\text{ (use $(a+b)(a-b)=a^2-b^2$)}
\\\\&=
\dfrac{4-9i^2}{13}
\\\\&=
\dfrac{4-9(-1)}{13}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{4+9}{13}
\\\\&=
\dfrac{13}{13}
\\\\&=
1
.\end{align*}
Since the product is equal to $1,$ then $
\dfrac{2+3i}{52}$ or $\dfrac{1}{26}+\dfrac{3}{52}i$
$ is the reciprocal of the given number.