To re-express \(3 - \sqrt{32}\) as a complex number, we first simplify \(\sqrt{32}\).
We note that: \[ \sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2} \]
Thus, we can rewrite the expression: \[ 3 - \sqrt{32} = 3 - 4\sqrt{2} \]
Since \(4\sqrt{2}\) is a positive number, \(3 - 4\sqrt{2}\) will yield a negative value. To express it in terms of the imaginary unit \(i = \sqrt{-1}\), we let: \[ 3 - 4\sqrt{2} = -|4\sqrt{2} - 3| \quad (\text{since } 4\sqrt{2} > 3) \]
Thus, we can express this in terms of imaginary numbers: \[ 3 - 4\sqrt{2} = -(4\sqrt{2} - 3) \] Therefore: \[ 3 - \sqrt{32} = -(4\sqrt{2} - 3) = -x \] where \(x = 4\sqrt{2} - 3\).
To express this as a complex number, we can write: \[ 3 - \sqrt{32} = 0 - \sqrt{(4\sqrt{2} - 3)^2} \cdot i \]
To finalize: \[ 3 - \sqrt{32} = -(4\sqrt{2} - 3)i \]
Thus, the complex number form is: \[ 3 - \sqrt{32} = -|4\sqrt{2} - 3| i \]
But substituting it back gives: \[ 3 - 4\sqrt{2} = -x \implies -x i \] Thus: \[ 3 - \sqrt{32} = -(4\sqrt{2} - 3)i \]
In conclusion: The re-expressed complex number form is: \[ 0 + (3 - \sqrt{32})i \] (it is possible with non real)
Therefore: \[ = 0 - (4\sqrt{2} - 3) i \]
So: \[ = 0 - (4\sqrt{2} - 3)i \] or \[ = -(4\sqrt{2} - 3)i \] Now \(3 - \sqrt{32} = a + bi\), it's perfectly clear you need to clearly derive imaginary constants for 0 and final coefficients.
This wants to purely confirm: The complex number representation can be summarized as \(a + bi\) = \(x + yi\).
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