How do you multiply $(8 - 2 i) (- 3 + 7 i)$ $(8-2i)(-3+7i)$ in trigonometric form?

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Answer 1 · 2016-05-08T04:49:23

FOIL

If it was $(8 - 2 x) (- 3 + 7 x)$ $(8-2x)(-3+7x)$

You would foil, and you would do that same thing.
Now, lets foil with $i$ $i$ .

First $(8 \cdot - 3) = - 24$ $(8*-3)=-24$
Outer $(8 \cdot 7 i) = 56 i$ $(8*7i) = 56i$
- Act $i$ $i$ like $x$ $x$
Inner $(- 2 i \cdot - 3) = 6 i$ $(-2i*-3)=6i$
Last $(- 2 i \cdot 7 i) = - 14 i^{2} = - 14 \cdot - 1 = 14$ $(-2i*7i)=-14i^2=-14*-1=14$
- Remember #i^2 = -1

Now put it all back together
$- 24 + 56 i + 6 i + 14$ $-24+56i+6i+14$

Combine like terms
$- 10 + 62 i$ $-10+62i$

How do you multiply (8−2i)(−3+7i) (8-2i)(-3+7i) in trigonometric form?