How do you divide #( 2i-6) / ( -2i +4 )# in trigonometric form?

1 Answer
Binayaka C.
Jul 25, 2016

In trigonometric form #Z=sqrt2[cos188.13+isin188.13]#

Explanation:

Let#Z=(2i-6)/(-2i+4)= ((2i-6)(4+2i))/((4+2i)(4-2i))= (4i^2-4i-24)/(16-4i^2) =(-4i-28)/20= -7/5-1/5i# [since #i^2=-1#] Modulus #Z=sqrt((-7/5)^2+(-1/5)^2) =sqrt2# Argument Z:#theta=tan^-1(1/5/7/5)=tan^-1(1/7)=8.13^0+180^0=188.13^0[180^0# is added as it is on 3rd quadrant] Hence in trigonometric form #Z=sqrt2[cos188.13+isin188.13]#[Ans]

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