Quivering thought

Given Equation $z^2+z+(1-a)=0$

Clearly, This equation will not have Real Roots if D<0 $1-4(1-a)<0$

$4a<3$

$a < \dfrac {3}{4}$

Therefore, a has to be lesser than $\dfrac {3}{4}$

Hence, $a$ cannot take the value $\dfrac {3}{4}$

Let $\displaystyle z = x + iy$ be the complex number.

Now, a can be written as:
$\displaystyle a = x^{2} + x + 2ixy +iy -y^{2} + 1$

For a to have real values, the imaginary part of a must be zero i.e. :

$\displaystyle 2ixy + iy = 0$

$=> iy(2x +1) = 0$ but

$i !=0$ and $y !=0$ because z has non-zero imaginary part, therefore, $2x+1 = 0$ or $\displaystyle x = -\frac{1}{2}$

Plugging this value in a gives: $\displaystyle a_{real} = \frac{1}{4} - \frac{1}{2} - y^{2} + 1$ or $\displaystyle a_{real} = \frac{3}{4} - y^2$ Given that $z$ has non zero imaginary part meaning that $y$ is never zero. So the value at $y=0$ is not possible for a. Hence, answer is $\boxed{\frac{3}{4}}$ .