Find the exact value of (sqrt3 + i)^6 and express it in trig form?

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find the exact value of #(sqrt3 + i)^6# and express it in trig form

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Answer 1 · 2018-01-02T13:06:55

In trigonometric form expressed as #64 (cos π + i sin π)#

#Z=a+ib #. Modulus: #|Z|=sqrt (a^2+b^2)#;

Argument:#theta=tan^-1(b/a)# Trigonometrical form :

#Z =|Z|(costheta+isintheta)#

#Z=(sqrt3+i)#. Modulus #|Z|=sqrt(( sqrt3 )^2+1^2)#

# =sqrt(3+1)=sqrt4=2#

Argument: #tan theta= 1/sqrt3:. theta=pi/6 #. Z lies on first

quadrant, so #theta =pi/6 :. Z=2(cos(pi/6)+isin(pi/6))#

De Moivre's Theorem: For every real number #theta# and every

positive integer # n, (cos θ + i sin θ)^n = cos nθ + i sin nθ #

#:.(sqrt3 + i)^6 = [2 (cos (π/6) + i sin (π/6))]^6#

#=64 (cos π + i sin π)#

In trigonometric form expressed as #64 (cos π + i sin π)# [Ans]