How do I find the trigonometric form of the complex number #3-3sqrt3 i#?

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Answer 1 · 2018-04-29T12:32:04

In trigonometric form: #6(cos 5.236+i sin 5.236) #

Let #Z=a+ib ; Z=3- 3 sqrt 3i ; a=3 ,b = - 3sqrt3# ;

#Z# is in #4# th quadrant. Modulus #|Z|=sqrt(a^2+b^2)#

#=(sqrt(3^2+ (-3 sqrt3)^2)) =sqrt 36 =6 #

# tan alpha =|b/a|= (3sqrt3)/3 or tan alpha =sqrt 3 #

#alpha = tan ^-1 (sqrt3) ~~ 1.0472#

#theta# is on #4#th quadrant # :. theta=2pi-1.0472~~ 5.236:.#

Argument , # theta =5.236 :. # In trigonometric form expressed as

#|Z|(cos theta+i sin theta) = 6(cos 5.236+i sin 5.236) # [Ans]