Answer
The standard form of the expression $\frac{6}{5+i}$ is $\frac{15}{13}-\frac{3}{13}i$.
Work Step by Step
Consider the expression,
$\frac{6}{5+i}$
The imaginary part is in the denominator, so we multiply the numerator and denominator by the complex conjugate of the denominator -- that is, for the complex number $\left( 5+i \right)$ its complex conjugate is $\left( 5-i \right)$.
Multiply the expression by $\frac{\left( 5-i \right)}{\left( 5-i \right)}$.
$\begin{align}
& \frac{6}{5+i}=\frac{6}{5+i}\cdot \frac{5-i}{5-i} \\
& =\frac{6\left( 5-i \right)}{\left( 5+i \right)\left( 5-i \right)}
\end{align}$
The product of the complex number $\left( a+bi \right)$ and its complex conjugate $\left( a-bi \right)$ results in a real number -- that is, $\left( a+bi \right)\left( a-bi \right)={{a}^{2}}+{{b}^{2}}$.
Therefore,
$\frac{6\left( 5-i \right)}{\left( 5+i \right)\left( 5-i \right)}=\frac{6\left( 5-i \right)}{{{5}^{2}}+{{1}^{2}}}$
Simplify the above expression
$\begin{align}
& \frac{6\left( 5-i \right)}{{{5}^{2}}+{{1}^{2}}}=\frac{30-6i}{25+1} \\
& =\frac{30-6i}{26}
\end{align}$
Further simplify the expression $\frac{30-6i}{26}$
$\begin{align}
& \frac{6\left( 5-i \right)}{{{5}^{2}}+{{1}^{2}}}=\frac{30-6i}{26} \\
& =\frac{2\left( 15-3i \right)}{2\left( 13 \right)} \\
& =\frac{15-3i}{13} \\
& =\frac{15}{13}-\frac{3}{13}i
\end{align}$
Hence, the standard form of the expression $\frac{6}{5+i}$ is $\frac{15}{13}-\frac{3}{13}i$.
