Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Appendix G - Exercises - Page A12: 18

Answer

a) $\overline{z+w}=\overline{z}+\overline{w}$ b) $\overline{zw}=\overline{z} \cdot \overline{w}$ c) $\overline{z^n}=\overline{z}^{n}$

Work Step by Step

a) Consider $z=a+bi; w=c+di$ $\overline{z+w}=\overline{(a+bi)+(c+di)}$ or, $\overline{z+w}=(a+c)-(b+d)i$ or, $\overline{z+w}=(a-bi)+(c-di)$ Thus, $\overline{z+w}=\overline{z}+\overline{w}$ b) $z=a+bi; w=c+di$ $\overline{zw}=\overline{(a+bi)(c+di)}$ or, $\overline{zw}=ac-bd-bci-adi$ or, $\overline{zw}=(a-bi)(c-di)$ Thus, $\overline{zw}=\overline{z} \cdot \overline{w}$ c) Apply De Movire's Theorem $\overline{z^n}=\overline{r^n(\cos n \theta+i \sin n \theta}$ or, $\overline{z^n}=[r(\cos \theta-i \sin \theta)]^n$ Thus, $\overline{z^n}=\overline{z}^{n}$
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