How do you simplify $\frac{1 - 2 i}{3 - 4 i}$ $(1-2i)/(3-4i)$ ?

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How do you simplify $\frac{1 - 2 i}{3 - 4 i}$ $(1-2i)/(3-4i)$ ?

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Answer 1 · 2018-03-09T21:09:08

$\frac{- 2 i + 11}{25}$ $(-2i+11)/25$

Explanation:

Multiply by the conjugate of $\frac{3 + 4 i}{3 + 4 i}$ $(3+4i)/(3+4i)$ to cancel out the imaginary numbers in the denominator. Also remember that $i^{2}$ $i^2$ is equal to -1.

$= \frac{1 - 2 i}{3 - 4 i} \cdot \frac{3 + 4 i}{3 + 4 i}$ $=(1-2i)/(3-4i) * (3+4i)/(3+4i)$
$= \frac{3 + 4 i - 6 i - 8 i^{2}}{9 + 12 i - 12 i - 16 i^{2}}$ $=(3+4i-6i-8i^2)/(9+12i-12i-16i^2)$
$= \frac{- 2 i + 3 - 8 (- 1)}{9 - 16 (- 1)}$ $=(-2i+3-8(-1))/(9-16(-1))$
$= \frac{- 2 i + 11}{25}$ $=(-2i+11)/25$

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How do you simplify $\frac{1 - 2 i}{3 - 4 i}$ $(1-2i)/(3-4i)$ ?

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