Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.4 - Graphs of Polar Equations - Exercise Set - Page 755: 101


The result in the standard form is $2$.

Work Step by Step

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 $2$.
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