How do you write the complex number in trigonometric form #3-3i#?

1 Answer
Dreamer.N
May 5, 2018

In the trigonometric form we will have: #3sqrt(2)(cos(-pi/4)+isin(-pi/4))#

Explanation:

We have
3-3i
Taking out 3 as common we have 3(1-i)
Now multiplying and diving by #sqrt2# we get, 3 #sqrt2#(1/ #sqrt2#- i/ #sqrt2#)

Now we have to find the argument of the given complex number which is tan(1/#sqrt2#/(-1/#sqrt2#)) whixh comes out to be -#pi#/4 .Since the sin part is negative but cos part is positive so it lies in quadrant 4, implying that argument is #-pi/4#.
Hence
#3sqrt(2)(cos(-pi/4)+isin(-pi/4))# is the answer.

Hope it helps!!

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