InterviewSolution
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InterviewSolution
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If `z=(sqrt3/2+i/2)^5+(sqrt3/2-i/2)^5`, thenA. `Re(z) =0`B. `Im ( z) =0`C. `Re(z) gt 0 ,Im(z) gt 0 `D. `Re(z) gt 0,Im (z) lt 0` |
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Answer» Correct Answer - B Given `z=(sqrt(3)/2+i/2)^5+(sqrt(3)/2-i/2)^5` `[because omega=(-1+isqrt(3))/(2) and omega^2=(-1-isqrt(3))/(2)]` Now , `sqrt(3+i)/2=-i((-1+isqrt(3))/2)=-iomega` and `(sqrt(3)-1)/2=((-1-isqrt(3))/2)=i omega^2` `therefore z=(-iomega)^5+(iomega^2)^5=iw^2+iw` `=i(omega-omega^2)=i(isqrt(3))=-sqrt(3)` `rArr Re(z) lt 0 andlm (z)=0` Alternate Solution We know that `z+bar z =2 Re (z)` If ` z=(sqrt(3)/2+i/2)^5 +(sqrt(3)/2-i/2)^5` then z is purely real ,i,e Im (z)=0 |
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