How do you simplify $\dfrac{1}{2+5i}$?

Hint: Try to simplify by doing rationalization. This can be done by multiplying $\left( 2-5i \right)$ both in numerator and in denominator. Then do the necessary calculations to obtain the required solution.

Complete step by step solution:
Rationalization: It is a method by which we can write a fraction in such a way that the denominator contains only rational numbers.
Considering our question $\dfrac{1}{2+5i}$
It can be rationalized by multiplying $\left( 2-5i \right)$ both in numerator and in denominator.
multiplying $\left( 2-5i \right)$ both in numerator and in denominator, we get
$\Rightarrow \dfrac{1\cdot \left( 2-5i \right)}{\left( 2+5i \right)\left( 2-5i \right)}$
For denominator part, as we know $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
So, applying the formula we get
$\left( 2+5i \right)\left( 2-5i \right)={{\left( 2 \right)}^{2}}-{{\left( 5i \right)}^{2}}=4-\left( -25 \right)=4+25=29$
Now our expression can be written as
$\Rightarrow \dfrac{2-5i}{29}$
Hence, the simplified form of $\dfrac{1}{2+5i}$ is $\dfrac{2-5i}{29}$.
This is the required solution of the given question.

Note: We know the value of $i=\sqrt{-1}$. So, ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$ which is used in the calculation of $\left( 2+5i \right)\left( 2-5i \right)$. The simplified form of $\dfrac{1}{2+5i}$ we obtained is a complex number and can be denoted as $Z=\dfrac{2-5i}{29}$. Again as we know for a complex number ‘Z’ it’s real part and imaginary part can be denoted as $\operatorname{Re}\left( Z \right)$ and $\operatorname{Im}\left( Z \right)$ respectively. Hence, for $Z=\dfrac{2-5i}{29}$, it’s real part is $\operatorname{Re}\left( Z \right)=\dfrac{2}{29}$ and the imaginary part is $\operatorname{Im}\left( Z \right)=-\dfrac{5}{29}$.