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

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Answer 1 · 2015-02-03T21:09:24

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

Let's go:

$(3+2i)(1-3i)=3*1+3*(-3i)+2i*1-2i*3i$
$=3-9i+2i-6i^2=3-7i+2i^2$

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

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

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