How do you add (2+8i)+(3+4i) in trigonometric form?

1 Answer
May 18, 2016

(2+8i)+(3+4i)=68[cosα+isinα]+5[cosβ+issinβ],

where α=arctan4 and β=arctan(43).

Explanation:

Let us first write (2+8i) and (3+4i) in trigonometric form.

a+ib can be written in trigonometric form rcosθa+irsinθ=r(cosθa+isinθ),
where r=a2+b2 and tanθ=ba or θ=arctan(ba)

Hence (2+8i)=22+82[cosα+isinα] or

68[cosα+isinα], where α=arctan4 and

(3+4i)=32+42[cosβ+isinβ] or

5[cosβ+issinβ], where β=arctan(43)

Hence (2+8i)+(3+4i) =

68[cosα+isinα]+5[cosβ+issinβ],

where α=arctan4 and β=arctan(43).