Question

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 4
A. `-1`
B. `1/3`
C. `1/2`
D. `3/4`

Answer 1

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`