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

1 Answer
Apr 29, 2018

Answer:

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

Explanation:

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]