InterviewSolution
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InterviewSolution
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Let z be a complex number such that the imaginary part of z is nonzero and a = z2 + z + 1 is real. Then a cannot take the value(A) –1 (B)1 3(C)1 2(D)3 4A. `-1`B. `1/3`C. `1/2`D. `3/4` |
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Answer» Correct Answer - D IF `ax^2+bx+c=0` has roots `alpha ,beta ` then `alpha,beta=(-bpmsqrt(b^2-4ac))/(2a)` For roots to be real `b^2-4ac ge0` ltbgt Description of situation As imeginary part of z=x+iy is non-zero `rArr y ne 0 ` Method I let z=x+iy `therefore " "a=(x+iy)^2+(x+iy)+1` `rArr (x^2-y^2+x+1-a)+i(2xy+y)=0` `rArr (x^2y^2+x+1-a)+iy(2x+1)=0` If is purely real if y(2x+1)=0 but imaginary part of z.ie y is non-zero `rArr 2x+1=0 or x=-1//2` From Eq.(i) `1/4-y^2=1/2+1a=0` `rArr a=-y^2+3/4 rArr a lt 3/4` Method II `Here z^2+z+(1-a)=0` `therefore " "z=(-1pmsqrt(1-4(1-a)))/(2xx1)` `rArr " "z=(-1pmsqrt(4a-3))/(2)` For z do not have real roots `4a-3 lt 0 rArr a lt 3/4` |
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