## Differential Equations and Linear Algebra (4th Edition)

Published by Pearson

# Appendix A - Review of Complex Numbers - Exercises for A - Problems - Page 795: 10

See below.

#### Work Step by Step

We know that $i^2=-1$ $z_1z_2=(1-2i)(3+4i)=3+4i-6i+8=11-2i$ $z_1/z_2=\frac{1-2i}{3+4i}=\frac{(1-2i)(3-4i)}{(3+4i)(3-4i)}=\frac{3-4i-6i-8}{9+16}=\frac{-5-10i}{25}$

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