Asked by Jackie
-8i(2i-7)
Answers
Answered by
Steve
All these complex value problems can be treated just like polynomials, where the reals and imaginaries have to be treated separately
-8i(2i) - 8i(-7)
-16i^2 + 56i
16 + 56i
Don't forget that i^2 = -1
also key: (a+bi)(a-bi) = a^2 - b^2i^2 = a^2+b^2
-8i(2i) - 8i(-7)
-16i^2 + 56i
16 + 56i
Don't forget that i^2 = -1
also key: (a+bi)(a-bi) = a^2 - b^2i^2 = a^2+b^2
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