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what is the product of the complex numbers (-3i+4) and (3i+4) ? A) 1 B) 7 C) 25 D) -7 + 24i E) 7 + 24i
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i times-ed the two together and got -9i + 16... maybe i need to add
then id get 8
use \(\large \bf (a+b)(a-b)=a^2-b^2\) property
\[\large \bf (4+3i)(4-3i)=4^2-(3i)^2\]
awesome! that helped a lot! at least i know were to start now. thanks!
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so far got... 16+12i+12-9i^2 = 16 -9i^2
correct !
so what is the value of \(\large \bf i^2\)?
eather (x-1) + (x+1) or add it with the other i^2
well \[\large \bf i=\sqrt{-1}\]
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so what would be \(\large \bf i^2\) then ?
√-9i^3 = -3 and 3
ops ^2
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