Answer
$\frac{2}{13}+\frac{3}{13}i$
Work Step by Step
$(2-3i)^{-1}=\displaystyle \frac{1}{2-3i}=\frac{1}{2-3i}*\frac{2+3i}{2+3i}=\frac{2+3i}{4-9i^{2}}=\frac{2+3i}{4-9*-1}=\frac{2+3i}{4+9}=\frac{2+3i}{13}=\frac{2}{13}+\frac{3}{13}i$
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter 1, Equations and Graphs - Section 1.5 - Complex Numbers - 1.5 Exercises - Page 131: 42
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