Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.3 The Product and Quotient Theorems - 8.3 Exercises - Page 369: 9

Answer

$-3i$

Work Step by Step

First, we use the product theorem to multiply the absolute values and add the arguments: $(\sqrt{3} cis 45^{\circ})(\sqrt{3} cis 225^{\circ}) \\=\sqrt{3} \sqrt{3} cis (45^{\circ}+225^{\circ}) \\=3 cis (270^{\circ})$ Next, we change the expression into its equivalent form: $=3 cis (270^{\circ}) \\=3(\cos270^{\circ}+i\sin270^{\circ})$ Since we know that $\cos270^{\circ} =0$ and $\sin270^{\circ}=-1$, we substitute these values in the expression and simplify: $3(\cos270^{\circ}+i\sin270^{\circ}) \\=3(-i) \\=-3i$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.