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If $z$ is a complex number having non zero imaginary part and $z^2+z+1=a$ where $a$ is real, then $a$ cannot take which of the following values?

(A) -1 (B) 1/2 (C) 1/3 (D) 3/4
Can you answer this question?
 
 

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Let $z=x+iy$ where $ y\neq 0$
given: $z^2+z+1=a$ where $a$ is real.
$\Rightarrow\:(x+iy)^2+(x+iy)+1=a$
$\Rightarrow\:(x^2-y^2+x+1)+i(2xy+y)=a$ (where $a$ is real)
$\therefore\:2xy+y=0$
$\Rightarrow\:x=\large-\frac{1}{2}$ since $\:y\neq 0$
$\therefore\:a=\large\frac{1}{4}$$-y^2-\large\frac{1}{2}$$+1$
$\Rightarrow\:a=\large\frac{3}{4}$$-y^2$
Since $y\neq 0$ $a$ cannot be $\large\frac{3}{4}$
answered Aug 5, 2013 by rvidyagovindarajan_1
 

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