What is the trigonometric form of # (-2-9i) #?

1 Answer
Topscooter
Dec 18, 2015

#sqrt(85)e^(iarccos(-2/85))#

Explanation:

You first need the module of this complex number, given by the formula #abs(-2-9i) = sqrt((-9)^2 + (-2)^2) = sqrt85#.

You can now factorize the complex number by its module : #-2-9i = sqrt(85)(-2/85 -i9/85)#. You can now say that #EEtheta in RR# such that #cos(theta) = -2/85# and #sin(theta) = -9/85#.

So #theta = arccos(-2/85)#.

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