Divide the following complex numbers. (-6+8i)/(-4-3i)

Multiply by conjugate: Multiply the numerator and denominator by the conjugate of the denominator: ((-6+8i)/(-4-3i)) * ((-4+3i)/(-4+3i))Multiply numerators: First, we'll multiply the numerators: (-6+8i) * (-4+3i) = (-6)(-4) + (-6)(3i) + (8i)(-4) + (8i)(3i) = 24 - 18i - 32i + 24i^2 Since i^2 = -1, we replace 24i^2 with 24(-1): = 24 - 18i - 32i - 24 = 24 - 24 - 18i - 32i = 0 - 50i = -50iReplace i^2: Now, we'll multiply the denominators: (-4-3i) * (-4+3i) = (-4)(-4) + (-4)(3i) - (3i)(-4) - (3i)(3i) = 16 - 12i + 12i - 9i^2 Again, replacing i^2 with -1: = 16 - 12i + 12i + 9 = 16 + 9 = 25Multiply denominators: Finally, we divide the results of the numerators by the results of the denominators: (-50i) / 25 = -2i

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Math Problems

Algebra 2

Add, subtract, multiply, and divide polynomials

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Q. Divide the following complex numbers.

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\frac{-6+8 i}{-4-3 i}

Multiply by conjugate: Multiply the numerator and denominator by the conjugate of the denominator: $\left(\frac{-6+8i}{-4-3i}\right) \cdot \left(\frac{-4+3i}{-4+3i}\right)$
Multiply numerators: First, we'll multiply the numerators: $\newline$ $(-6+8i) \times (-4+3i) = (-6)(-4) + (-6)(3i) + (8i)(-4) + (8i)(3i)$ $\newline$ $= 24 - 18i - 32i + 24i^2$ $\newline$ Since $i^2 = -1$ , we replace $24i^2$ with $24(-1)$ : $\newline$ $= 24 - 18i - 32i - 24$ $\newline$ $= 24 - 24 - 18i - 32i$ $\newline$ $= 0 - 50i$ $\newline$ $= -50i$
Replace $i^2$ : Now, we'll multiply the denominators: $\newline$ $(-4-3i) * (-4+3i) = (-4)(-4) + (-4)(3i) - (3i)(-4) - (3i)(3i)$ $\newline$ $= 16 - 12i + 12i - 9i^2$ $\newline$ Again, replacing $i^2$ with $-1$ : $\newline$ $= 16 - 12i + 12i + 9$ $\newline$ $= 16 + 9$ $\newline$ $= 25$
Multiply denominators: Finally, we divide the results of the numerators by the results of the denominators: $(-50i) / 25 = -2i$