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MathematicsComplex NumberCube Root of UnityHard2 minPYQ_2024
MathematicsHardsingle choice

Ifzis a complex number, then the number of common roots of the equationz1985+z100+1=0andz3+2z2+2z+1=0, is equal to :

Options:

Answer:
B
Solution:

Given,

z1985+z100+1=0  &  z3+2z2+2z+1=0

Now solving,

z3+2z2+2z+1=0

z+1z2-z+1+2zz+1=0

z+1z2+z+1=0

z=-1, z=ω, ω2

Now putting  z=ω in z1985+z100+1 we get,

ω1985+ω100+1

ω2+ω+1=0 as ω3n=1

Also, z=ω2

ω3970+ω200+1

ω+ω2+1=0

Also, z=-1 will not satisfy the equation z1985+z100+1

Hence, there are two common root

Stream:JEESubject:MathematicsTopic:Complex NumberSubtopic:Cube Root of Unity
2mℹ️ Source: PYQ_2024

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