Precalculus (6th Edition) Blitzer

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

Chapter 6 - Section 6.5 - Complex Numbers in Polar Form; DeMoivre's Theorem - Exercise Set - Page 769: 110


See the explanation below.

Work Step by Step

Apply the division operation on the provided two complex numbers. $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)}{{{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)}$ Multiply the numerator and denominator of the above equation with the conjugate of the denominator’s second factor. $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)}{{{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)}\cdot \frac{\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}{\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}$ …… (1) Apply the multiplication operation in equation (1). $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}\left[ \cos {{\theta }_{1}}\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)+i\sin {{\theta }_{1}}\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right) \right]}{{{r}_{2}}\left[ \cos {{\theta }_{2}}\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)+i\sin {{\theta }_{2}}\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right) \right]}$ $=\frac{{{r}_{1}}\left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-i\left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+i\left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-{{i}^{2}}\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right]}{{{r}_{2}}\left[ \left( \cos {{\theta }_{2}}\cdot \cos {{\theta }_{2}} \right)-i\left( \cos {{\theta }_{2}}\cdot \sin {{\theta }_{2}} \right)+i\left( \sin {{\theta }_{2}}\cdot \cos {{\theta }_{2}} \right)-{{i}^{2}}\left( \sin {{\theta }_{2}}\cdot \sin {{\theta }_{2}} \right) \right]}$ …… (2) Substitute the value of the iota in equation (2) and then rearrange the equation. $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}\left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+i\left\{ \left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-\left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right\} \right]}{{{r}_{2}}\left[ \left( \cos {{\theta }_{2}}\cdot \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{2}}\cdot \sin {{\theta }_{2}} \right)+i\left\{ \left( \sin {{\theta }_{2}}\cdot \cos {{\theta }_{2}} \right)-\left( \cos {{\theta }_{2}}\cdot \sin {{\theta }_{2}} \right) \right\} \right]}$ Further solve it to get $=\frac{{{r}_{1}}\left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+i\left\{ \left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-\left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right\} \right]}{{{r}_{2}}\left[ \left( {{\cos }^{2}}{{\theta }_{2}}+{{\sin }^{2}}{{\theta }_{2}} \right)+i\left( 0 \right) \right]}$ …… (3) Apply the identity of trigonometry ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ in the denominator of equation (3). $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}\left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+i\left\{ \left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-\left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right\} \right]}{{{r}_{2}}\left[ 1+i\left( 0 \right) \right]}$ Further solve it to get $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\cdot \left[ \left( \cos {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right)+i\left\{ \left( \sin {{\theta }_{1}}\cdot \cos {{\theta }_{2}} \right)-\left( \cos {{\theta }_{1}}\cdot \sin {{\theta }_{2}} \right) \right\} \right]$ …… (4) Apply the difference formula of cosine and sine in equation (4). $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\cdot \left[ \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right]$ Hence, the quotient of the provided two complex numbers is $\frac{{{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)}{{{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)}=\frac{{{r}_{1}}}{{{r}_{2}}}\cdot \left[ \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right]$.
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