Question

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

Answer 1

`(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]