How do you multiply $(4 - 8 i) (3 + 9 i)$ $(4-8i)(3+9i)$ in trigonometric form?

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Answer 1 · 2018-08-06T07:13:32

$(4 - 8 i) (3 + 9 i) = 12 + 36 i - 24 i - 72 i^{2}$ $(4-8i)(3+9i)=12+36i-24i-72i^2$

= $12 + 12 i - (- 72)$ $12+12i-(-72)$ because $i^{2} = - 1$ $i^2=-1$

= $84 + 12 i$ $84+12i$

How do you multiply (4−8i)(3+9i) (4-8i)(3+9i) in trigonometric form?