## Precalculus (6th Edition) Blitzer

The result in the standard form is $2$.
Perform the standard operation and expand for the given expression $\frac{\left( 2+2i \right)}{\left( 1+i \right)}$. By multiplying the expression $\frac{\left( 2+2i \right)}{\left( 1+i \right)}$ with $\frac{\left( 1-i \right)}{\left( 1-i \right)}$ we get, \begin{align} & =\frac{\left( 2+2i \right)}{\left( 1+i \right)}\times \frac{\left( 1-i \right)}{\left( 1-i \right)} \\ & =\frac{\left( 2+2i \right)\left( 1-i \right)}{\left( 1+i \right)\left( 1-i \right)} \\ & =\frac{2-2i+2i-2{{i}^{2}}}{1-i+i-{{i}^{2}}} \\ & =\frac{2+2}{1+1}=2 \\ \end{align} Therefore, the result in the standard form is $2$.