## Precalculus (6th Edition) Blitzer

The value of the expression $\frac{1+i}{1+2i}+\frac{1-i}{1-2i}$ in the standard form is$\frac{6}{5}+0i$.
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$\frac{6}{5}+0i$.