Answer
$\dfrac{2}{29}-\dfrac{5}{29}i$
Work Step by Step
The reciprocal of the given complex number, $
2+5i
,$ is
\begin{align*}
&
\dfrac{1}{2+5i}
.\end{align*}
Multiplying the numerator and the denominator by the conjugate of the denominator, the expression above is equivalent to
\begin{align*}
&
\dfrac{1}{2+5i}\cdot\dfrac{2-5i}{2-5i}
\\\\&=
\dfrac{2-5i}{(2+5i)(2-5i)}
.\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{2-5i}{(2)^2-(5i)^2}
\\\\&=
\dfrac{2-5i}{4-25i^2}
\\\\&=
\dfrac{2-5i}{4-25(-1)}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{2-5i}{4+25}
\\\\&=
\dfrac{2-5i}{29}
.\end{align*}
The reciprocal is $
\dfrac{2-5i}{29}
$.
Checking:
Multiplying the given expression, $
2+5i
$ and the reciprocal, $
\dfrac{2-5i}{29}
,$ results to
\begin{align*}
&
(2+5i)\cdot\left( \dfrac{2-5i}{29} \right)
\\\\&=
\dfrac{(2+5i)(2-5i)}{29}
\\\\&=
\dfrac{(2)^2-(5i)^2}{29}
&\text{ (use $(a+b)(a-b)=a^2-b^2$)}
\\\\&=
\dfrac{4-25i^2}{29}
\\\\&=
\dfrac{4-25(-1)}{29}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{4+25}{29}
\\\\&=
\dfrac{29}{29}
\\\\&=
1
.\end{align*}
Since the product is equal to $1,$ then $
\dfrac{2-5i}{29}$ or $\dfrac{2}{29}-\dfrac{5}{29}i$
$ is the reciprocal of the given number.