Let's start by considering [tex]\( u = -4i \)[/tex]. We need to fill in the missing number so that [tex]\( u = -4i \)[/tex] satisfies the equation:
[tex]\[ u^2 + \square = 0 \][/tex]
First, we calculate [tex]\( u^2 \)[/tex]:
[tex]\[ u = -4i \][/tex]
[tex]\[ u^2 = (-4i)^2 \][/tex]
Now, using the property of imaginary numbers where [tex]\( i^2 = -1 \)[/tex]:
[tex]\[ (-4i)^2 = (-4)^2 \cdot (i^2) \][/tex]
[tex]\[ = 16 \cdot (-1) \][/tex]
[tex]\[ = -16 \][/tex]
So, [tex]\( u^2 = -16 \)[/tex]。
In order for [tex]\( u^2 + \square = 0 \)[/tex] to hold true, we need:
[tex]\[ -16 + \square = 0 \][/tex]
Solving for the missing number:
[tex]\[ \square = 16 \][/tex]
Thus, the equation becomes:
[tex]\[ u^2 + 16 = 0 \][/tex]
Now, let's solve for [tex]\( u \)[/tex] in the equation [tex]\( u^2 + 16 = 0 \)[/tex]:
[tex]\[ u^2 = -16 \][/tex]
[tex]\[ u = \pm \sqrt{-16} \][/tex]
We know that the square root of a negative number involves the imaginary unit [tex]\( i \)[/tex], with [tex]\( \sqrt{-16} = 4i \)[/tex]:
[tex]\[ u = \pm 4i \][/tex]
So, the solutions are:
[tex]\[ u = 4i \quad \text{and} \quad u = -4i \][/tex]
In summary, the missing number to fill is 16, and the two solutions to the equation [tex]\( u^2 + 16 = 0 \)[/tex] are:
[tex]\[ u = -4i \quad \text{and} \quad u = 4i \][/tex]
