1. 6(cos 30° + i sin 30°) =
2. 8(cos 270° + i sin 270°) =
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To solve these problems, we'll use Euler's formula, which states that for any real number x:
cos x + i sin x = e^(ix)
And to find the powers of complex numbers in polar form, we use De Moivre's theorem:
(r(cos x + i sin x))^n = r^n(cos(nx) + i sin(nx))
Now let's solve the given problems:
1. 6(cos 30° + i sin 30°) =
Since we don't have any powers here, we can simplify the expression directly:
6[cos(30°) + i sin(30°)]
6[(√3/2) + i(1/2)]
= 3√3 + 3i
So the expression equals 3√3 + 3i.
2. 8(cos 270° + i sin 270°) =
Again, we don't have any powers, so we can simplify the expression directly:
8[cos(270°) + i sin(270°)]
8[(0) + i(-1)]
= -8i
So the expression equals -8i.