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

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

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Answer 1 · 2018-05-02T22:48:17

$\frac{4 + 2 i}{- 1 + i} ∣ \cdot (- 1 - i)$ $(4+2i)/(-1+i) | *(-1-i)$

$\frac{(4 + 2 i) (- 1 - i)}{(- 1 + i) (- 1 - i)}$ $((4+2i)(-1-i))/((-1+i)(-1-i))$

$\frac{- 2 i^{2} - 6 i - 4}{1 - i^{2}}$ $(-2i^2-6i-4)/(1-i^2)$

$\frac{2 - 6 i - 4}{1 + 1}$ $(2-6i-4)/(1+1)$

$\frac{- 2 - 6 i}{2}$ $(-2-6i)/(2)$

$= - 1 - 3 i$ $=-1-3i$

Explanation:

We want to get rid of $i$ $i$ in the the bottom of the fraction in order to get it on Certesian form. We can do this by multiplying with $(- 1 - i)$ $(-1-i)$ .

This will give us,

$\frac{(4 + 2 i) (- 1 - i)}{(- 1 + i) (- 1 - i)}$ $((4+2i)(-1-i))/((-1+i)(-1-i))$

$\frac{- 2 i^{2} - 6 i - 4}{1 - i^{2}}$ $(-2i^2-6i-4)/(1-i^2)$

Out from here we know that $i^{2} = - 1$ $i^2=-1$ and $- i^{2} = 1$ $-i^2=1$ . So we can get rid of the $i^{2}$ $i^2$ too. Leaving us to

$\frac{- 2 - 6 i}{2}$ $(-2-6i)/(2)$

$= - 1 - 3 i$ $=-1-3i$

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

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How do you simplify (4+ 2i) /( -1 + i)4+2i−1+i(4+ 2i) /( -1 + i)?

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