Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.1 - Complex Numbers - Exercise Set - Page 315: 53


The standard form of the expression ${{\left( 2+i \right)}^{2}}-{{\left( 3-i \right)}^{2}}$ is $-5+10i$.

Work Step by Step

Consider the expression,${{\left( 2+i \right)}^{2}}-{{\left( 3-i \right)}^{2}}$ Use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$. \[\begin{align} & {{\left( 2+i \right)}^{2}}-{{\left( 3-i \right)}^{2}}=\left( {{2}^{2}}+2\cdot 2\cdot i+{{i}^{2}} \right)-\left( {{3}^{2}}-2\cdot 3\cdot i+{{i}^{2}} \right) \\ & =\left( 4+4i+{{i}^{2}} \right)-\left( 9-6i+{{i}^{2}} \right) \\ & =4+4i+{{i}^{2}}-9+6i-{{i}^{2}} \end{align}\] Combine real and imaginary terms. \[\begin{align} & {{\left( 2+i \right)}^{2}}-{{\left( 3-i \right)}^{2}}=4-9+4i+6i+{{i}^{2}}-{{i}^{2}} \\ & =-5+10i \end{align}\] Therefore, the standard form of the expression ${{\left( 2+i \right)}^{2}}-{{\left( 3-i \right)}^{2}}$ is $-5+10i$.
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