Answer
The value of the expression $\frac{1+i}{1+2i}+\frac{1-i}{1-2i}$ in the standard form is\[\frac{6}{5}+0i\].
Work Step by Step
Consider the expression,
$\frac{1+i}{1+2i}+\frac{1-i}{1-2i}$
Take the least common denominator, that is $\left( 1+2i \right)\left( 1-2i \right)$.
Multiply the expression $\frac{1+i}{1+2i}$ by \[\frac{\left( 1-2i \right)}{\left( 1-2i \right)}\] and the $\frac{1-i}{1-2i}$by\[\frac{\left( 1+2i \right)}{\left( 1+2i \right)}\].
$\begin{align}
& \frac{1+i}{1+2i}+\frac{1-i}{1-2i}=\frac{1+i}{1+2i}\cdot \frac{\left( 1-2i \right)}{\left( 1-2i \right)}+\frac{1-i}{1-2i}\cdot \frac{\left( 1+2i \right)}{\left( 1+2i \right)} \\
& =\frac{\left( 1+i \right)\left( 1-2i \right)+\left( 1-i \right)\left( 1+2i \right)}{\left( 1+2i \right)\left( 1-2i \right)}
\end{align}$
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 $\left( a+bi \right)\left( a-bi \right)={{a}^{2}}+{{b}^{2}}$.
Apply it,
$\begin{align}
& \frac{1+i}{1+2i}+\frac{1-i}{1-2i}=\frac{\left( 1+i \right)\left( 1-2i \right)+\left( 1-i \right)\left( 1+2i \right)}{\left( 1+2i \right)\left( 1-2i \right)} \\
& =\frac{\left( 1+i \right)\left( 1-2i \right)+\left( 1-i \right)\left( 1+2i \right)}{{{1}^{2}}+{{2}^{2}}} \\
& =\frac{\left( 1+i \right)\left( 1-2i \right)+\left( 1-i \right)\left( 1+2i \right)}{1+4} \\
& =\frac{\left( 1+i \right)\left( 1-2i \right)+\left( 1-i \right)\left( 1+2i \right)}{5}
\end{align}$
The foil method: if an expression has two binomials multiplied together then, for any arbitrary numbers w, x, y and z,
$\left( w+x \right)\left( y+z \right)=wy+wz+xy+xz$
Apply it,
\[\begin{align}
& \frac{1+i}{1+2i}+\frac{1-i}{1-2i}=\frac{\left( 1+i \right)\left( 1-2i \right)+\left( 1-i \right)\left( 1+2i \right)}{5} \\
& =\frac{1-2i+i-2{{i}^{2}}+1+2i-i-2{{i}^{2}}}{5} \\
& =\frac{2-4{{i}^{2}}}{5}
\end{align}\]
Recall that,
$\begin{align}
& \sqrt{-1}=i \\
& -1={{i}^{2}}
\end{align}$
Apply it,
\[\begin{align}
& \frac{1+i}{1+2i}+\frac{1-i}{1-2i}=\frac{2-4{{i}^{2}}}{5} \\
& =\frac{2-4\left( -1 \right)}{5} \\
& =\frac{2+4}{5} \\
& =\frac{6}{5}
\end{align}\]
Write it in the standard form $a+bi$.
\[\begin{align}
& \frac{1+i}{1+2i}+\frac{1-i}{1-2i}=\frac{2-4{{i}^{2}}}{5} \\
& =\frac{2-4\left( -1 \right)}{5} \\
& =\frac{2+4}{5} \\
& =\frac{6}{5}+0i
\end{align}\]
Thus, the value of the expression $\frac{1+i}{1+2i}+\frac{1-i}{1-2i}$ in the standard form is\[\frac{6}{5}+0i\].
