# How do you multiply  (-1-6i)(4-i)  in trigonometric form?

Mar 28, 2017

Multiplication of two numbers in trigonometric form:

${r}_{1} \left(\cos \left({\theta}_{1}\right) + i \sin \left({\theta}_{1}\right)\right) \times {r}_{2} \left(\cos \left({\theta}_{2}\right) + i \sin \left({\theta}_{2}\right)\right) = {r}_{1} {r}_{2} \left(\cos \left({\theta}_{1} + {\theta}_{2}\right) + i \sin \left({\theta}_{1} + {\theta}_{2}\right)\right)$

#### Explanation:

For the given numbers:

${r}_{1} = \sqrt{- {1}^{2} + - {6}^{2}}$

${r}_{1} = \sqrt{37}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{- 6}{-} 1\right) + \pi$

${\theta}_{1} = {\tan}^{-} 1 \left(6\right) + \pi$

Please notice that we add $\pi$ because the signs of "a" and "b" tell us that the angle is in the 3rd quadrant.

${r}_{2} = \sqrt{{4}^{2} + - {1}^{2}}$

${r}_{2} = \sqrt{17}$

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

Please notice that we add $2 \pi$ because the signs of "a" and "b" tell us that the angle is in the 4th quadrant.

Do the multiplication:

${r}_{1} \left(\cos \left({\theta}_{1}\right) + i \sin \left({\theta}_{1}\right)\right) \times \sqrt{17} \left(\cos \left({\theta}_{2}\right) + i \sin \left({\theta}_{2}\right)\right) = {r}_{1} {r}_{2} \left(\cos \left({\theta}_{1} + {\theta}_{2}\right) + i \sin \left({\theta}_{1} + {\theta}_{2}\right)\right)$

$\sqrt{37} \left(\cos \left({\tan}^{-} 1 \left(6\right) + \pi\right) + i \sin \left({\tan}^{-} 1 \left(6\right) + \pi\right)\right) \times \sqrt{17} \left(\cos \left({\tan}^{-} 1 \left(\frac{- 1}{4}\right) + 2 \pi\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 1}{4}\right) + 2 \pi\right)\right) = \sqrt{37} \sqrt{17} \left(\cos \left({\tan}^{-} 1 \left(6\right) + \pi + {\tan}^{-} 1 \left(\frac{- 1}{4}\right) + 2 \pi\right) + i \sin \left({\tan}^{-} 1 \left(6\right) + \pi + {\tan}^{-} 1 \left(\frac{- 1}{4}\right) + 2 \pi\right)\right)$