How do you multiply  (3-5i)(7-8i)  in trigonometric form?

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Nov 13, 2017

$\left(3 - 5 i\right) \left(7 - 8 i\right) = 61.98 \left(\cos 4.40 + i \sin 4.40\right)$

Explanation:

$Z = \left(3 - 5 i\right) \left(7 - 8 i\right) = \left(21 - 24 i - 35 i + 40 {i}^{2}\right)$

$= 21 - 59 i - 40 = - 19 - 59 i \left[{i}^{2} = - 1\right]$

Modulus $| Z | = r = \sqrt{{\left(- 19\right)}^{2} + {\left(- 59\right)}^{2}} = 61.98$ ;

$\tan \alpha = \frac{b}{a} = \frac{- 59}{-} 19 = 3.105 \therefore \alpha = {\tan}^{-} 1 \left(3.105\right) = 1.259$ or

$\theta$ is on $3 r d$ quadrant $\therefore \theta = 1.259 + \pi = 4.40$ rad

Argument : $\theta = 4.40 \therefore$ In trigonometric form expressed as

$r \left(\cos \theta + i \sin \theta\right) = 61.98 \left(\cos 4.40 + i \sin 4.40\right) \therefore$

$\left(3 - 5 i\right) \left(7 - 8 i\right) = 61.98 \left(\cos 4.40 + i \sin 4.40\right)$ [Ans]

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