Let z be a complex number such that the imaginary part of z is nonzero and a = z^2 + z + 1 is real. Then a cannot take the value

Options

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

Explanation

To solve the problem, we need to find the values that the expression \( a = z^2 + z + 1 \) can take, given that \( z \) is a complex number with a non-zero imaginary part, and \( a \) is real.

### Step-by-Step Solution:

1. **Express \( z \) in terms of its real and imaginary parts:**
Let \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part, with \( y \neq 0 \).

2. **Calculate \( z^2 \):**
\[ z^2 = (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + (2xy)i \]

3. **Substitute \( z \) and \( z^2 \) into the expression for \( a \):**
\[ a = z^2 + z + 1 = (x^2 - y^2 + 2xyi) + (x + iy) + 1 \]
\[ = (x^2 - y^2 + x + 1) + (2xy + y)i \]

4. **Separate the real and imaginary parts:**
The real part of \( a \) is \( x^2 - y^2 + x + 1 \) and the imaginary part is \( 2xy + y \).

5. **Set the imaginary part to zero (since \( a \) is real):**
\[ 2xy + y = 0 \]
Factor out \( y \):
\[ y(2x + 1) = 0 \]

6. **Analyze the equation:**
Since \( y \neq 0 \), we must have:
\[ 2x + 1 = 0 \implies x = -\frac{1}{2} \]

7. **Substitute \( x = -\frac{1}{2} \) back into the expression for \( a \):**
\[ a = \left(-\frac{1}{2}\right)^2 - y^2 - \frac{1}{2} + 1 \]
\[ = \frac{1}{4} - y^2 - \frac{1}{2} + 1 \]
\[ = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} - y^2 \]
\[ = \frac{3}{4} - y^2 \]

8. **Determine the values \( a \) can take:**
Since \( y^2 \) is always positive (as \( y \neq 0 \)), \( a \) can take values less than \( \frac{3}{4} \).

9. **Conclusion:**
Therefore, \( a \) cannot take the value \( \frac{3}{4} \).

### Final Answer:
The value that \( a \) cannot take is \( \frac{3}{4} \).