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

May 14, 2018

(-7+ 3 i)(1+ 3 i) =24.08 ( cos (3.99) + i sin(3.99) = (-16-18 i)

#### Explanation:

Let Z=a+i b ; Z=-7+ 3 i ; a=-7 ,b = 3 ;

$Z = - 7 + 3 i$ is in $2$ nd quadrant.

Modulus $| Z | = \sqrt{{a}^{2} + {b}^{2}} = \left(\sqrt{{\left(- 7\right)}^{2} + {3}^{2}}\right) = \sqrt{58}$

$\tan \alpha = | \frac{b}{a} | = \frac{3}{7} \mathmr{and} \alpha = {\tan}^{-} 1 \left(\frac{3}{7}\right) \approx 0.4049$

$\theta$ is on $2$ nd quadrant $\therefore \theta = \pi - 0.4049$

$\therefore \theta \approx 2.7367$. Argument , $\theta \approx 2.7367 \therefore$

In trigonometric form expressed as

$r \left(\cos \theta + i \sin \theta\right) = \sqrt{58} \left(\cos 2.74 + i \sin 2.74\right)$

$Z = 1 + 3 i$ is in $1$ st quadrant.

Modulus $| Z | = \sqrt{{a}^{2} + {b}^{2}} = \left(\sqrt{{1}^{2} + {3}^{2}}\right) = \sqrt{10}$

$\tan \alpha = | \frac{b}{a} | = \frac{3}{1} \mathmr{and} \alpha = {\tan}^{-} 1 \left(3\right) \approx 1.249$

$\theta$ is on $1$ st quadrant $\therefore \theta = 1.249$

Argument , $\theta \approx 1.249 \therefore$

In trigonometric form expressed as

$r \left(\cos \theta + i \sin \theta\right) = \sqrt{10} \left(\cos 1.25 + i \sin 1.25\right)$

$\left(- 7 + 3 i\right) \left(1 + 3 i\right) =$

$\sqrt{58} \left(\cos 2.74 + i \sin 2.74\right) \cdot \sqrt{10} \left(\cos 1.25 + i \sin 1.25\right) \approx$

sqrt58 * sqrt 10 ( cos (2.74+1.25) + i sin(2.74+1.25) ~~

24.08 ( cos (3.99) + i sin(3.99) = (-16-18 i) [Ans]