1
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Question

# Let z be a complex number such that the imaginary part of z is non - zero and a=z2+z+1 is real. Then, a cannot take the value

A
1
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B
13
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C
12
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D
34
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Solution

## The correct option is D 34If ax2+bx+c=0 has roots α,β then α,β=−b±√b2−4ac2a For roots to be b2−4ac≥0. Description of Situation As imaginary part of z = x+iy is non- zero. ⇒y≠0 Method 1: Let z=x+iy ∴a=(x+iy)2+(x+iy)+1 ⇒(x2−y2+x+1−a)+i(2xy+y)=0⇒(x2−y2+x+1−a)+iy(2x+1)=0, ...(1) It is purely real, if y(2x+1) = 0 But imaginary part of z. i.e y is non-zero. ⇒2x+1=0orx=−12 From Eq. (1), 14−y2−12+1−a=0 ⇒a=−y2+34⇒a<34 Method 2: Here, z2+z+(1−a)=0 ∴z=−1±√1−4(1−α)2⇒z=−1±√(4a)−32 For z do not have real roots, 4a−3<0⇒a<34

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