Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a=z^2+z+1$ is real. Then $a$ cannot take the value
Step-by-Step Solution
Key Concept: Setting $\text{Im}(a)=0$ with $\text{Im}(z)\neq0$ fixes $\text{Re}(z)=-1/2$, making $a=3/4-y^2<3/4$ strictly. The upper bound $3/4$ is never achieved.
**Step 1: Force Im(a)=0 with Im(z)≠0**
Let $z=x+iy$, $y\neq0$. Then $a=(x^2-y^2+x+1)+iy(2x+1)$. For $a$ real: $y(2x+1)=0$. Since $y\neq0$, we need $x=-\dfrac{1}{2}$.
**Step 2: Find the range of a**
With $x=-\dfrac{1}{2}$: $a=\dfrac{1}{4}-y^2-\dfrac{1}{2}+1=\dfrac{3}{4}-y^2$. Since $y\neq0$, $y^2>0$, so $a<\dfrac{3}{4}$.
**Step 3: Identify the excluded value**
All of $-1,\,\dfrac{1}{3},\,\dfrac{1}{2}$ are $<\dfrac{3}{4}$ and achievable. But $a=\dfrac{3}{4}$ requires $y=0$, contradicting $\text{Im}(z)\neq0$.
Correct Answer: 4