• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q. 39

Expert-verified
Found in: Page 592

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Find zw and $\frac{z}{w}$. Leave the answers in polar form. role="math" localid="1646661399572" $z=2+2i\phantom{\rule{0ex}{0ex}}w=\sqrt{3}-i$

zw in polar form is $4\sqrt{2}\left(\mathrm{cos}15°+i\mathrm{sin}15°\right)$.

$\frac{z}{w}$ in polar form is $\sqrt{2}\left(\mathrm{cos}75°+i\mathrm{sin}75°\right)$.

See the step by step solution

## Step 1. Given information.

Consider the given question,

$z=2+2i\phantom{\rule{0ex}{0ex}}w=\sqrt{3}-i$

The multiplication of the polar form of complex numbers is given by,

${z}_{1}{z}_{2}={r}_{1}{r}_{2}\left[\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)\right]......\left(i\right)\phantom{\rule{0ex}{0ex}}\frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\left[\mathrm{cos}\left({\theta }_{1}-{\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{1}-{\theta }_{2}\right)\right]......\left(ii\right)$

We need to write the given numbers in polar form.

$r=\sqrt{{x}^{2}+{y}^{2}},\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$

## Step 2. Find the polar form of w.

Determine the polar form of z,

${r}_{z}=\sqrt{{2}^{2}+{2}^{2}}\phantom{\rule{0ex}{0ex}}{r}_{z}=2\sqrt{2}\phantom{\rule{0ex}{0ex}}{\theta }_{z}={\mathrm{tan}}^{-1}\left(\frac{2}{2}\right)\phantom{\rule{0ex}{0ex}}{\theta }_{z}=45°$

Therefore, $z=2\sqrt{2}\left(\mathrm{cos}45°+i\mathrm{sin}45°\right)$.

Determine the polar form of w,

${r}_{w}=\sqrt{{\left(\sqrt{3}\right)}^{2}+{\left(-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}{r}_{w}=2\phantom{\rule{0ex}{0ex}}{\theta }_{w}={\mathrm{tan}}^{-1}\left(\frac{-1}{\sqrt{3}}\right)\phantom{\rule{0ex}{0ex}}{\theta }_{w}=-30°$

Therefore, $w=2\left(\mathrm{cos}\left(-30°\right)+i\mathrm{sin}\left(-30°\right)\right)$

## Step 3. Find zw and zw.

Substitute the values of z,w in equation (i),

$z·w=2\sqrt{2}\left[\mathrm{cos}\left(45°+i\mathrm{sin}45°\right)\right]·2\left[\mathrm{cos}\left(-30°\right)+i\mathrm{sin}\left(-30°\right)\right]\phantom{\rule{0ex}{0ex}}z·w=\left(2\right)2\sqrt{2}\left[\mathrm{cos}\left(45°+\left(-30°\right)\right)+i\mathrm{sin}\left(45°+\left(-30°\right)\right)\right]\phantom{\rule{0ex}{0ex}}z·w=4\sqrt{2}\left(\mathrm{cos}15°+i\mathrm{sin}15°\right)$

Substitute the values of z,w in equation (ii),

$\frac{z}{w}=\frac{2\sqrt{2}\left(\mathrm{cos}45°+i\mathrm{sin}45°\right)}{2\left(\mathrm{cos}\left(-30°\right)+i\mathrm{sin}\left(-30°\right)\right)}\phantom{\rule{0ex}{0ex}}\frac{z}{w}=\frac{2\sqrt{2}}{2}\left[\mathrm{cos}\left(45°-\left(-30°\right)\right)+i\mathrm{sin}\left(45°-\left(-30°\right)\right)\right]\phantom{\rule{0ex}{0ex}}\frac{z}{w}=\sqrt{2}\left(\mathrm{cos}75°+i\mathrm{sin}75°\right)$