How do you multiply $(3 + 2 i) (1 - 3 i)$ $(3+2i)(1-3i)$ ?

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How do you multiply $(3 + 2 i) (1 - 3 i)$ $(3+2i)(1-3i)$ ?

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Answer 1 · 2015-02-03T21:09:24

Just as any other polynomial, except whenever you get $i^{2}$ $i^2$ you change it to $- 1$ $-1$

Let's go:

$(3 + 2 i) (1 - 3 i) = 3 \cdot 1 + 3 \cdot (- 3 i) + 2 i \cdot 1 - 2 i \cdot 3 i$ $(3+2i)(1-3i)=3*1+3*(-3i)+2i*1-2i*3i$
$= 3 - 9 i + 2 i - 6 i^{2} = 3 - 7 i + 2 i^{2}$ $=3-9i+2i-6i^2=3-7i+2i^2$

The $i^{2}$ $i^2$ changes into $- 1$ $-1$ so the answer is:

$3 - 7 i + 2 \cdot (- 1) = 3 + 7 i - 2 = 1 + 7 i$ $3-7i+2*(-1)=3+7i-2=1+7i$

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How do you multiply $(3 + 2 i) (1 - 3 i)$ $(3+2i)(1-3i)$ ?

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