Answer
$\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}i$
Work Step by Step
The reciprocal of the given complex number, $
a+bi
,$ is
\begin{align*}\require{cancel}
&
\dfrac{1}{a+bi}
.\end{align*}
Multiplying the numerator and the denominator by the conjugate of the denominator, the expression above is equivalent to
\begin{align*}
&
\dfrac{1}{a+bi}\cdot\dfrac{a-bi}{a-bi}
\\\\&=
\dfrac{a-bi}{(a+bi)(a-bi)}
.\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{a-bi}{(a)^2-(bi)^2}
\\\\&=
\dfrac{a-bi}{a^2-b^2i^2}
\\\\&=
\dfrac{a-bi}{a^2-b^2(-1)}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{a-bi}{a^2+b^2}
.\end{align*}
The reciprocal is $
\dfrac{a-bi}{a^2+b^2}
$.
Checking:
Multiplying the given expression, $
a+bi
$ and the reciprocal, $
\dfrac{a-bi}{a^2+b^2}
,$ results to
\begin{align*}
&
(a+bi)\cdot\left( \dfrac{a-bi}{a^2+b^2}
\right)
\\\\&=
\dfrac{(a+bi)(a-bi)}{a^2+b^2}
\\\\&=
\dfrac{(a)^2-(bi)^2}{a^2+b^2}
&\text{ (use $(a+b)(a-b)=a^2-b^2$)}
\\\\&=
\dfrac{a^2-b^2i^2}{a^2+b^2}
\\\\&=
\dfrac{a^2-b^2(-1)}{a^2+b^2}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{a^2+b^2}{a^2+b^2}
\\\\&=
1
.\end{align*}
Since the product is equal to $1,$ then $
\dfrac{a-bi}{a^2+b^2}$ or $\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}i
$ is the reciprocal of the given number.
