Answer
The value of the operation for the expression $\frac{8}{1+\frac{2}{i}}$ in the standard form is\[\frac{8}{5}+\frac{16}{5}i\]
Work Step by Step
Consider the expression,
$\frac{8}{1+\frac{2}{i}}$
Multiply the expression $1$ by $\frac{i}{i}$ .
$\begin{align}
& \frac{8}{1+\frac{2}{i}}=\frac{8}{1\cdot \frac{i}{i}+\frac{2}{i}} \\
& =\frac{8}{\frac{i+2}{i}} \\
& =\frac{8}{\frac{i+2}{\frac{i}{1}}} \\
& =\frac{8}{i+2}\cdot \frac{i}{1}
\end{align}$
Simplify it further as,
$\begin{align}
& \frac{8}{1+\frac{2}{i}}=\frac{8i}{i+2} \\
& =\frac{8i}{2+i}
\end{align}$
Multiply the denominator term by its complex conjugate; that is, multiply the term $\frac{8i}{2+i}$ by $\frac{2-i}{2-i}$.
$\begin{align}
& \frac{8}{1+\frac{2}{i}}=\frac{8i}{i+2} \\
& =\frac{8i}{2+i}\cdot \frac{2-i}{2-i} \\
& =\frac{8i\left( 2-i \right)}{\left( 2-i \right)\left( 2+i \right)}
\end{align}$
The product of a complex number $\left( a+bi \right)$ and its complex conjugate $\left( a-bi \right)$ results in a real number, that is $\left( a+bi \right)\left( a-bi \right)={{a}^{2}}+{{b}^{2}}$.
Apply it,
$\begin{align}
& \frac{8}{1+\frac{2}{i}}=\frac{8i\left( 2-i \right)}{\left( 2-i \right)\left( 2+i \right)} \\
& =\frac{8i\left( 2-i \right)}{{{2}^{2}}+{{1}^{2}}} \\
& =\frac{8i\left( 2-i \right)}{4+1} \\
& =\frac{8i\left( 2-i \right)}{5}
\end{align}$
The distributive property states that, $a\cdot \left( b+c \right)=a\cdot b+a\cdot c$.
Apply it,
$\begin{align}
& \frac{8}{1+\frac{2}{i}}=\frac{8i\left( 2-i \right)}{5} \\
& =\frac{16i-8{{i}^{2}}}{5}
\end{align}$
Recall that,
$\begin{align}
& \sqrt{-1}=i \\
& -1={{i}^{2}}
\end{align}$
Apply it,
$\begin{align}
& \frac{8}{1+\frac{2}{i}}=\frac{8i\left( 2-i \right)}{5} \\
& =\frac{16i-8{{i}^{2}}}{5} \\
& =\frac{16i-8\left( -1 \right)}{5} \\
& =\frac{16i+8}{5}
\end{align}$
Write it in the standard form $a+bi$.
$\begin{align}
& \frac{8}{1+\frac{2}{i}}=\frac{16i+8}{5} \\
& =\frac{8}{5}+\frac{16}{5}i
\end{align}$
Thus, the value of the operation for the expression $\frac{8}{1+\frac{2}{i}}$ in the standard form is $\frac{8}{5}+\frac{16}{5}i$.
