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

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Answer 1 · 2017-12-30T13:02:47

$- 7 + 4 ι$ $-7+4iota$

$ι = \sqrt{- 1}$ $iota= sqrt(-1)$
So, $ι^{2} = - 1$ $iota^2=-1$

$(- 1 + 2 ι) (3 + 2 ι)$ $(-1+2iota) (3+2iota)$

$- 3 - 2 ι + 6 ι - 4$ $-3-2iota+6iota-4$

$- 7 + 4 ι$ $-7+4iota$

Simply multiply as we generally multiply polynomial by polynomial provided the rules of complex must in mind just like $ι^{2} = - 1$ $iota^2=-1$ .

How do you multiply (−1+2i)(3+2i) (-1+2i)(3+2i) in trigonometric form?