Please Help With This Trig Problem

Question

5(cos 220° + i sin 220°)/2(cos 115° + i sin 115°)

A. 5/2(cos 105° + i sin 105°)

B. 2/5(cos 105° + i sin 105°)

C. 5/2(cos 335° + i sin 335°)

D. 10(cos 335° + i sin 335°)

Answer 1

apply the rule; simply divide the constants and subtract the angles

5(cos220°+isin220°)/2(cos115°+isin115°)

(5/2)[(cos220-115)+isin(220-115)]

(5/2)(cos105°+isin105°) which is A.

Answer 2

Remember Euler's formula, e^(ix) = cos(x) + i*sin(x). So then we have:

5(cos(220) + i*sin(220))/(2(cos(115)+i*sin(115))) = 5e^(i*220)/2e^(i*115) = 5/2*e^(i*105) = 5/2(cos(105) + i*sin(105)).

Note: technically when using Euler's formula you're supposed to use radians instead of degrees, but it doesn't change the answer in this case because you're transforming it back.