If `z in C, z !in R, and a = z^(2) + 3z + 5`, then a cannot take value

Options

A. `-2//5`
B. `5//2`
C. `11/4`
D. `-11/5`

Answer

To solve the problem, we need to analyze the expression \( a = z^2 + 3z + 5 \) where \( z \) is a complex number that is not a real number. We want to find out what value \( a \) cannot take.

### Step-by-Step Solution:

**Step 1: Rewrite the expression in a different form.**
We start with the expression:
\[ a = z^2 + 3z + 5 \]
To analyze this expression, we can complete the square for the quadratic part.

**Step 2: Complete the square.**
We can rewrite \( z^2 + 3z \) as follows:
\[ z^2 + 3z = \left(z + \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 \]
This gives us:
\[ z^2 + 3z = \left(z + \frac{3}{2}\right)^2 - \frac{9}{4} \]
Now substituting this back into the expression for \( a \):
\[ a = \left(z + \frac{3}{2}\right)^2 - \frac{9}{4} + 5 \]

**Step 3: Simplify the expression.**
We can simplify the constant terms:
\[ 5 - \frac{9}{4} = \frac{20}{4} - \frac{9}{4} = \frac{11}{4} \]
Thus, we have:
\[ a = \left(z + \frac{3}{2}\right)^2 + \frac{11}{4} \]

**Step 4: Analyze the expression.**
Since \( z \) is a complex number and not a real number, \( z + \frac{3}{2} \) is also not a real number. The square of a complex number can take any non-negative value, but since \( z \) is not real, \( \left(z + \frac{3}{2}\right)^2 \) will not be zero.

**Step 5: Determine the value of \( a \).**
The minimum value of \( \left(z + \frac{3}{2}\right)^2 \) is \( 0 \) (which occurs if \( z + \frac{3}{2} = 0 \)), but since \( z \) cannot be real, this minimum value cannot be achieved. Therefore, \( \left(z + \frac{3}{2}\right)^2 \) must be a positive number.

Thus, the minimum value of \( a \) is:
\[ a_{\text{min}} = 0 + \frac{11}{4} = \frac{11}{4} \]

**Conclusion:**
Since \( a \) can never be equal to \( \frac{11}{4} \) (as that would imply \( z \) is real), we conclude that \( a \) cannot take the value:
\[ \boxed{\frac{11}{4}} \]