How do you multiply # (15-3i)(4+7i) # in trigonometric form?

1 Answer
Binayaka C.
Apr 11, 2017

#(15-3i)(4+7i) = 123.33 (cos 0.854 + i sin 0.854) = 81 + 93i #

Explanation:

#15-3i ; r=sqrt(15^2+3^2)=sqrt234, theta=2*pi-tan^-1 (3/15)=6.086 #

#:. 15-3i = sqrt 234 (cos 6.086 + isin 6.086)#

#4+7i ; r=sqrt(4^2+7^2)=sqrt65, theta=tan^-1 (7/4)=1.052 #

#:. 4+7i = sqrt 65 (cos 1.052 + isin 1.052)#

#(15-3i)(4+7i) = sqrt234*sqrt65(cos (6.086+1.052-2*pi) + isin(6.086+1.052-2*pi) = 123.33 (cos 0.854 + i sin 0.854) = 81+93i#

#(15-3i)(4+7i) = 123.33 (cos 0.854 + i sin 0.854) = 81 + 93i # [Ans]

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