Answer:
Option B
Option C
Step-by-step explanation:
[tex]$ i = \sqrt{-1} $[/tex]
⇒ [tex]$ i^2 = (\sqrt{-1})^2 = -1 $[/tex], which is a real number.
Similarly, [tex]$ i^3 = i^2.i = -1 \times i = -i $[/tex]
And we also know that [tex]$ i^4 = 1 $[/tex]
So, we see that odd powers of [tex]$ i $[/tex] is a complex number and even powers renders us a real number.
Using this we solve the problem.
Option A: [tex]$ (-4i)^{11} $[/tex] 11 is an odd power. This would be a complex number and not a real number.
Option B: [tex]$ (-3i)^{12} $[/tex] Even power of i. So, this should give us a real number.
Option C: [tex]$ (2 + 3i)^2 $[/tex]
(a + bi)² = a² - b² + 2abi
So, (2 + 3i)² would be a complex number because of the 2bi term.
Option D: (4 + 5i)(4 - 5i)
(a + ib)(a - ib) = a² + b²
So, (4 + 5i)(4 - 5i) = 4² + 5² = 41, a real number.
Option E: (6 + 8i)(8 + 6i)
(a + ib)(c + id) = ac - bd + (ad + bc)i
This would be a complex number because of the (ad + bc) term.
So, Options B and D are real numbers.
