How do you multiply $(1 + 7 i) (- 3 + 2 i)$ $(1+7i)(-3+2i)$ in trigonometric form?

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Answer 1 · 2017-11-19T11:05:42

$(1 + 7 i) (- 3 + 2 i) \approx 25.495 (cos 228.18 + i sin 228.18)$ $(1+7i)(-3+2i) ~~ 25.495 (cos 228.18+isin228.18)$

$Z = (1 + 7 i) (- 3 + 2 i) = (- 3 + 2 i - 21 i + 14 i^{2})$ $Z= (1+7i)(-3+2i) = (-3+2i-21i+14i^2)$

$= - 3 - 19 i - 14 = - 17 - 19 i [i^{2} = - 1]$ $=-3-19i-14 = -17-19i [i^2= -1]$

$Z$ $Z$ lies on $3 r d$ $3rd$ quadrant

Modulus $| Z | = r = \sqrt{{(- 17)}^{2} + {(- 19)}^{2}} \approx 25.495$ $|Z|=r=sqrt((-17)^2+ (-19)^2) ~~25.495$ ;

$tan alpha =b/a= (-19)/-17 ~~ 1.1176 :. alpha =tan^-1(1.1176) ~~ 48.18^0$

Since $Z$ is on $3rd$ quadrant $:. theta=pi+alpha$ or

$theta= 180+48.18= 228.18^0$ . Argument : $theta =228.18^0$

In trigonometric form expressed as $r(cos theta+isintheta)$

$= 25.495 (cos 228.18+isin228.18)$

$:. (1+7i)(-3+2i) ~~ 25.495 (cos 228.18+isin228.18)$ [Ans]

How do you multiply (1+7i)(-3+2i) (1+7i)(−3+2i) (1+7i)(-3+2i) in trigonometric form?