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Answers without the blur. Sign up and see all textbooks for free! Q. 42

Expert-verified Found in: Page 592 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # Write the expression in the standard form $a+ib$.${\left[3\left(\mathrm{cos}80°+i\mathrm{sin}80°\right)\right]}^{3}$

The expression in the standard form is $-\frac{27}{2}-\frac{27\sqrt{3}}{2}i$.

See the step by step solution

## Step 1. Given information.

Consider the given question,

${\left[3\left(\mathrm{cos}80°+i\mathrm{sin}80°\right)\right]}^{3}$

We know that for a complex number raise to n when it is in the polar form,

${z}^{n}={r}^{n}\left(\mathrm{cos}\theta +i\mathrm{sin}\left(n\theta \right)\right)......\left(i\right)$

Write the expression in the standard form $z=\left(a+bi\right)$,

${z}^{3}={\left[3\left(\mathrm{cos}80°+i\mathrm{sin}80°\right)\right]}^{3}$

## Step 2. Use equation (i) and solve the equation.

Using the equation (i),

${z}^{3}={3}^{3}\left[\mathrm{cos}\left(3·80°\right)+i\mathrm{sin}\left(3·80°\right)\right]\phantom{\rule{0ex}{0ex}}{z}^{3}=27\left(\mathrm{cos}240°+i\mathrm{sin}240°\right)\phantom{\rule{0ex}{0ex}}{z}^{3}=27\left[-\frac{1}{2}+i\left(\frac{-\sqrt{3}}{2}\right)\right]\phantom{\rule{0ex}{0ex}}{z}^{3}=-\frac{27}{2}-\frac{27\sqrt{3}}{2}i$ ### Want to see more solutions like these? 