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Question

If z2z+1=0, then possible value(s) of znzn, where n is even number

A
3i
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B
0
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C
3i
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D
2
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Solution

The correct option is C 3i
z2z+1=0z=1±i32
z=ω or z=ω2

If z=ω, then
znzn=(ω)n(ω)n
=(1)nωn(1)nωn
=ωnω2n
Here, there are three cases possible
Case 1:n is multiple of 3
=ωnω2n=11=0
Case 2:n=3m+1,mZ+
=ω3m+1ω2(3m+1)=ω3mωω6mω2
=ωω2=3i
Case 3:n=3m+2,mZ+
=ω3m+2ω2(3m+2)=ω3mω2ω6mω4
=ω2ω=3i

If z=ω2, then
znzn=(ω2)n(ω2)n
=(1)nω2n(1)nω2n
=ω2nωn
Similarly as above we can do.

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