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

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Answer 1 · 2017-10-17T12:02:47

$(3 - 5 i) (7 - 8 i) = 61.98 (cos 4.40 + i sin 4.40)$ $(3-5i)(7-8i) = 61.98(cos 4.40+isin4.40)$

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

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

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

$tan alpha =b/a= (-59)/-19 = 3.105 :. alpha =tan^-1(3.105) = 1.259$ or

$theta$ is on $3rd$ quadrant $:. theta=1.259+pi= 4.40$ rad

Argument : $theta =4.40 :.$ In trigonometric form expressed as

$r(cos theta+isintheta) = 61.98(cos 4.40+isin4.40) :.$

$(3-5i)(7-8i) = 61.98(cos 4.40+isin4.40)$ [Ans]

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