Let z be a complex number such that the imaginary part of z is nonzero and a = z2 + z + 1 is real. Then a cannot take the value (A) –1 (B) 1 3 (C) 1 2 (D) 3 4

Options

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

Explanation

To solve the problem, we need to analyze the expression \( a = z^2 + z + 1 \) under the condition that \( a \) is real and the imaginary part of \( z \) is non-zero.

### Step 1: Write \( z \) 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 \( y \neq 0 \)).

### Step 2: Substitute \( z \) into the expression for \( a \)
Substituting \( z \) into the expression for \( a \):
\[ a = (x + iy)^2 + (x + iy) + 1 \]
Expanding this:
\[ a = (x^2 - y^2 + 2xyi) + (x + iy) + 1 \]
Combining like terms:
\[ a = (x^2 - y^2 + x + 1) + (2xy + y)i \]

### Step 3: Separate the real and imaginary parts
The real part of \( a \) is \( x^2 - y^2 + x + 1 \) and the imaginary part is \( (2xy + y) \).

### Step 4: Set the imaginary part to zero
Since \( a \) is real, the imaginary part must equal zero:
\[ 2xy + y = 0 \]
Factoring out \( y \) (since \( y \neq 0 \)):
\[ y(2x + 1) = 0 \]
This implies:
\[ 2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2} \]

### Step 5: Substitute \( x \) back into the expression for \( a \)
Now substitute \( x = -\frac{1}{2} \) into the real part of \( a \):
\[ a = \left(-\frac{1}{2}\right)^2 - y^2 - \frac{1}{2} + 1 \]
Calculating this:
\[ a = \frac{1}{4} - y^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} - y^2 \]

### Step 6: Analyze the possible values of \( a \)
Since \( y^2 \) is always positive (as \( y \neq 0 \)), \( a \) can be expressed as:
\[ a = \frac{3}{4} - y^2 \]
This means \( a \) can take values less than or equal to \( \frac{3}{4} \) but cannot equal \( \frac{3}{4} \) itself.

### Conclusion
Thus, \( a \) cannot take the value \( \frac{3}{4} \).

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

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