How do you add #(8+8i)+(-4+6i)# in trigonometric form?

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Answer 1 · 2017-10-07T15:19:27

See below.

To convert complex numbers to trigonometric form, find #r#, the distance of the point away from the origin, and #theta#, the angle.

#(8 + 8i) + (-4 + 6i) = 8 - 4 + 8i + 6i = 4 + 14i#

#4 + 14i# is in the form #a+bi#. First, find #r#:

#r^2 = a^2 + b^2#

#r^2 = 4^2 + 14^2#

#r = sqrt252 = 2sqrt53#

Find #theta#:

#tan theta = b/a#

#tan theta = 14/4#

#theta = tan^-1(14/4) ~~ 1.29#

In trigonometric form, this is #r(cos theta + i sin theta)# or in shorthand,
#r# #cis# #theta#.

Thus the answer is #2sqrt53 (cos 1.29 + i sin 1.29)# or #2sqrt 53# #cis# #1.29#.