Let z be a complex number such that `Im(z) ne 0. "If a" = z^(2) + 5z + 7`, then a cannot take value ____________

Answer

To solve the problem, we need to analyze the expression \( a = z^2 + 5z + 7 \) where \( z \) is a complex number with a non-zero imaginary part. We want to determine the value that \( a \) cannot take.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression for \( a \) is given as:
\[ a = z^2 + 5z + 7 \]

2. **Substituting \( z \)**:
Let \( z = x + yi \), where \( x \) and \( y \) are real numbers, and \( y \neq 0 \) (since the imaginary part of \( z \) is not zero).

3. **Calculating \( z^2 \)**:
We calculate \( z^2 \):
\[ z^2 = (x + yi)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + (2xy)i \]

4. **Calculating \( 5z \)**:
Now calculate \( 5z \):
\[ 5z = 5(x + yi) = 5x + 5yi \]

5. **Combining the Terms**:
Now substitute \( z^2 \) and \( 5z \) back into the expression for \( a \):
\[ a = (x^2 - y^2 + 5x + 7) + (2xy + 5y)i \]
This can be separated into real and imaginary parts:
\[ a = (x^2 - y^2 + 5x + 7) + (2xy + 5y)i \]

6. **Identifying the Imaginary Part**:
The imaginary part of \( a \) is \( 2xy + 5y \). Since \( y \neq 0 \), we can factor this expression:
\[ 2xy + 5y = y(2x + 5) \]
For \( a \) to be purely real, the imaginary part must equal zero:
\[ y(2x + 5) = 0 \]
Since \( y \neq 0 \), we must have:
\[ 2x + 5 = 0 \implies x = -\frac{5}{2} \]

7. **Substituting \( x \) Back**:
Now, substitute \( x = -\frac{5}{2} \) back into the expression for the real part of \( a \):
\[ a = \left(-\frac{5}{2}\right)^2 - y^2 + 5\left(-\frac{5}{2}\right) + 7 \]
Simplifying this:
\[ = \frac{25}{4} - y^2 - \frac{25}{2} + 7 \]
Converting \( -\frac{25}{2} \) to quarters:
\[ = \frac{25}{4} - y^2 - \frac{50}{4} + \frac{28}{4} \]
\[ = \frac{25 - 50 + 28 - 4y^2}{4} = \frac{3 - 4y^2}{4} \]

8. **Finding the Value of \( a \)**:
The expression \( \frac{3 - 4y^2}{4} \) shows that as \( y \) varies, \( a \) will take different values. However, for \( a \) to be real, \( 3 - 4y^2 \) must be non-negative:
\[ 3 - 4y^2 \geq 0 \implies y^2 \leq \frac{3}{4} \]
Therefore, the maximum value of \( a \) occurs when \( y = 0 \), which is not allowed since \( y \neq 0 \).

9. **Conclusion**:
The value \( a \) cannot be purely real, specifically it cannot take the value \( \frac{3}{4} \) when \( y^2 \) is maximized. Thus, the value that \( a \) cannot take is:
\[ \boxed{\frac{3}{4}} \]