How do you multiply $(- 1 + 9 i) (3 - 2 i)$ $(-1+9i)(3-2i)$ in trigonometric form?

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Answer 1 · 2018-06-25T19:00:15

$(- 1 + 9 i) (3 - 2 i) = - 3 + 2 i + 27 i - 18 i^{2}$ $(-1+9i)(3-2i)=-3+2i+27i-18i^2$

$- 3 + 29 i + 18$ $-3+29i+18$

$15 + 29 i$ $15+29i$

How do you multiply (−1+9i)(3−2i) (-1+9i)(3-2i) in trigonometric form?