To solve for the modulus [tex]\(|z|\)[/tex], we need to carefully analyze the given binary operation [tex]\(\otimes\)[/tex] and the equation involving [tex]\(z\)[/tex].
Given:
[tex]\[ u \otimes v = ac + bdi, \][/tex]
we need to find [tex]\(z\)[/tex] such that:
[tex]\[ z \otimes z = z^2 + 40. \][/tex]
First, we express [tex]\(z\)[/tex] in terms of its real and imaginary parts:
[tex]\[ z = a + bi, \][/tex]
where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are real numbers.
Applying the binary operation [tex]\(\otimes\)[/tex] to [tex]\(z\)[/tex]:
[tex]\[ z \otimes z = (a + bi) \otimes (a + bi). \][/tex]
Using the definition of [tex]\(\otimes\)[/tex]:
[tex]\[ z \otimes z = a \cdot a + b \cdot b \cdot i = a^2 + b^2 i. \][/tex]
Next, we need to evaluate [tex]\(z^2 + 40\)[/tex]. Squaring [tex]\(z\)[/tex] yields:
[tex]\[ (a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 + 2abi - b^2. \][/tex]
Then adding 40:
[tex]\[ z^2 + 40 = (a^2 - b^2) + 2abi + 40. \][/tex]
According to the problem statement:
[tex]\[ z \otimes z = z^2 + 40. \][/tex]
Equating the two expressions, we get:
[tex]\[ a^2 + b^2 i = (a^2 - b^2 + 40) + 2ab i. \][/tex]
Equating the real and imaginary parts:
[tex]\[ \text{Real part: } a^2 = a^2 - b^2 + 40, \][/tex]
[tex]\[ \text{Imaginary part: } b^2 = 2ab. \][/tex]
From the real part equation:
[tex]\[ a^2 = a^2 - b^2 + 40 \Rightarrow b^2 = 40. \][/tex]
The imaginary part suggests:
[tex]\[ b^2 = 2ab. \][/tex]
Given [tex]\( b^2 = 40 \)[/tex], substituting this into the imaginary equation yields:
[tex]\[ 40 = 2ab \Rightarrow ab = 20. \][/tex]
Now we have:
[tex]\[ b^2 = 40, \][/tex]
[tex]\[ ab = 20. \][/tex]
We can solve for [tex]\(a\)[/tex] by dividing the second equation by [tex]\(b\)[/tex]:
[tex]\[ a = \frac{20}{b}. \][/tex]
Substituting [tex]\( b^2 = 40 \)[/tex]:
[tex]\[ b = \pm \sqrt{40} = \pm 2\sqrt{10}. \][/tex]
For [tex]\(b = 2\sqrt{10}\)[/tex]:
[tex]\[ a = \frac{20}{2\sqrt{10}} = \frac{20}{2\sqrt{10}} = \sqrt{10}. \][/tex]
For [tex]\(b = -2\sqrt{10}\)[/tex]:
[tex]\[ a = \frac{20}{-2\sqrt{10}} = -\sqrt{10}. \][/tex]
Thus, the potential values for [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are:
[tex]\[ (a, b) = (\sqrt{10}, 2\sqrt{10}) \text{ or } (-\sqrt{10}, -2\sqrt{10}). \][/tex]
Finally, calculating the modulus [tex]\( |z| \)[/tex]:
[tex]\[ |z| = \sqrt{a^2 + b^2}. \][/tex]
Substituting [tex]\(a = \sqrt{10}\)[/tex] and [tex]\(b = 2\sqrt{10}\)[/tex]:
[tex]\[ |z| = \sqrt{(\sqrt{10})^2 + (2\sqrt{10})^2} = \sqrt{10 + 4 \cdot 10} = \sqrt{10 + 40} = \sqrt{50} = 5\sqrt{2}. \][/tex]
Therefore, [tex]\( |z| \)[/tex] is:
[tex]\[ \boxed{5\sqrt{2}}. \][/tex]
