Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Appendix A - Review of Complex Numbers - Exercises for A - Problems - Page 795: 7


See below.

Work Step by Step

We know that $i^2=-1$ $z_1z_2=(-1+3i)(2-i)=-2+i+6i+3=1+7i$ $z_1/z_2=\frac{-1+3i}{2-i}=\frac{(-1+3i)(2+i)}{(2-i)(2+i)}=\frac{-2-i+6i-3}{4+1}=\frac{-5+5i}{5}=-1+i$
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