What is the trigonometric form of # (2-4i)*(3-2i) #?

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Answer 1 · 2017-07-03T17:07:34

See the explanation below.

First, expand the expression.

#(2-4i) * (3-2i)#
= #6 - 4i - 12i +8i^2#
= #6 - 4i - 12i -8#
= #-2 - 16i#

To convert this to trigonometric form, you need to know the values of #r# and #theta#.

You can use the following equations:
#r^2 = x^2+y^2# and #tan theta = (y)/(x)#

#r^2 = x^2+y^2#
#r = sqrt(x^2+y^2)#
#r = sqrt((-2)^2+(-16)^2)#
#r = 2sqrt65#

#tan theta = (y)/(x)#
#tan theta = (-16)/(-2)#
#tan theta = 8#
#theta = tan^-1(8)#
#theta ~~ 1.45#

So, the answer is #2sqrt65# #cis# #1.45# or #2sqrt65 (cos 1.45 + i sin 1.45)#.