Answer
See below.
Work Step by Step
We know that $i^2=-1$
$z_1z_2=(2+3i)(1-i)=2-2i+3i+3=5+i$
$z_1/z_2=\frac{2+3i}{1-i}=\frac{(2+3i)(1+i)}{(1-i)(1+i)}=\frac{2+2i+3i-3}{1+1}=\frac{-1+5i}{2}=-1/2+5/2i$
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
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: 8
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