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How will we work with binomials containing radicals and complex numbers?
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sounds like a philosophy question you got an actual example?
(4+3i)(4-3i) 16+9 25
ok that looks good
\((a+bi)(a-bi)=x^2+b^2\) always
OOPS i means \(a^2+b^2\)
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What the super smart guy is sayin O.O
if you multiply out \[(a+bi)(a-bi)\] you get \[a^2-abi+abi-bi^2\] the middle terms add up to zero, and since \(i^2=-1\) the last term is \(+b^2\) giving a final answer of \[a^2+b^2\]
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