# How do you multiply  (-1-7i)(-3-4i)  in trigonometric form?

Feb 23, 2018

Thus,
$\left(- 1 - 7 i\right) \left(- 3 - 4 i\right) = 10 \sqrt{2} c i s \frac{7 \pi}{4}$

#### Explanation:

$\text{Let,} {z}_{1} = - 1 - 7 i ,$

$R e \left({z}_{1}\right) = - 1 , \text{ } I m \left({z}_{1}\right) = - 7$

${r}_{1} = \sqrt{{\left(- 1\right)}^{2} + {\left(- 7\right)}^{2}} = \sqrt{50} = 5 \sqrt{2}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{- 7}{- 1}\right) = \pi + {\tan}^{-} 1 \left(7\right)$

$\text{Let,} {z}_{2} = - 3 - 4 i$

$R e \left({z}_{2}\right) = - 3 , \text{ } I m \left({z}_{2}\right) = - 4$

${r}_{2} = \sqrt{{\left(- 3\right)}^{2} + {\left(- 4\right)}^{2}} = \sqrt{25} = 5$

${\theta}_{2} = {\tan}^{-} 1 \left(\frac{- 4}{- 3}\right) = \pi + {\tan}^{-} 1 \left(\frac{4}{3}\right)$

${z}_{1} = {r}_{1} c i s {\theta}_{1}$
${z}_{2} = {r}_{2} c i s {\theta}_{2}$

${z}_{1} {z}_{2} = \left({r}_{1} c i s {\theta}_{1}\right) \left({r}_{2} c i s {\theta}_{2}\right)$

${z}_{1} {z}_{2} = {r}_{1} {r}_{2} c i s {\theta}_{1} c i s {\theta}_{2}$
By De-Moivre's theorem

$c i s {\theta}_{1} c i s {\theta}_{2} = c i s \left({\theta}_{1} + {\theta}_{2}\right)$

Thus,

${z}_{1} {z}_{2} = {r}_{1} {r}_{2} c i s \left({\theta}_{1} + {\theta}_{2}\right)$

Substituting,

${r}_{1} {r}_{2} = 5 \sqrt{2} \times 5 = 10 \sqrt{2}$
${\theta}_{1} + {\theta}_{2} = \pi + {\tan}^{-} 1 \left(7\right) + \pi + {\tan}^{-} 1 \left(\frac{4}{3}\right)$
$= 2 \pi + {\tan}^{-} 1 \left(7\right) + {\tan}^{-} 1 \left(\frac{4}{3}\right)$

${\tan}^{-} 1 \left(7\right) + {\tan}^{-} 1 \left(\frac{4}{3}\right) = {\tan}^{-} 1 \left(\frac{7 + \frac{4}{3}}{1 - 7 \times \frac{4}{3}}\right)$

$\frac{7 + \frac{4}{3}}{1 - 7 \times \frac{4}{3}} = \frac{7 \times 3 + 4}{3 - 7 \times 4} = \frac{21 + 4}{3 - 28} = \frac{25}{-} 25 = \frac{1}{-} 1$
${\tan}^{-} 1 \left(7\right) + {\tan}^{-} 1 \left(\frac{4}{3}\right) = {\tan}^{-} 1 \left(\frac{1}{-} 1\right) = 2 \pi - {\tan}^{-} 1 \left(1\right)$

${\tan}^{-} 1 \left(1\right) = \frac{\pi}{4}$

$2 \pi - {\tan}^{-} 1 \left(1\right) = 2 \pi - \frac{\pi}{4} = \frac{7 \pi}{4}$

${r}_{1} c i s {\theta}_{1} {r}_{2} c i s {\theta}_{2} = 10 \sqrt{2} c i s \frac{7 \pi}{4}$

Thus,
$\left(- 1 - 7 i\right) \left(- 3 - 4 i\right) = 10 \sqrt{2} c i s \frac{7 \pi}{4}$