How do you write the complex number in trigonometric form #5/2(sqrt3-i)#?

1 Answer
Binayaka C.
Dec 20, 2016

In trigonometric form expressed as #5(cos330+isin330)#

Explanation:

#Z=a+ib #. Modulus: #|Z|=sqrt (a^2+b^2)#; Argument:#theta=tan^-1(b/a)# Trigonometrical form : #Z =|Z|(costheta+isintheta)#
#Z=5/2(sqrt3-i)= 5/2sqrt3 -5/2i #. Modulus #|Z|=sqrt(( 5/2sqrt3 )^2+(-5/2)^2) =sqrt(75/4+25/4)=sqrt25=5#
Argument: #tan theta= (-cancel(5/2))/(cancel(5/2)sqrt3)= -1/sqrt3 #. Z lies on fourth quadrant, so #theta =tan^-1(-1/sqrt3) = -pi/6=-30^0 or theta = 360-30=330^0 :. Z=5(cos330+isin330)#
In trigonometric form expressed as #5(cos330+isin330)#[Ans]

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