blessbails
Simplify the expression.
(–2 – 2i)(–4 + 6i)
>8 – 4i
>8 – 12i
>20 – 4i
>–4 – 4i



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SheldonEinstein
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Do you know the value for i (iota) ?

surdawi
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just multiply everything and you should get the answer (i*i=1)

UnkleRhaukus
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\[\quad(–2 – 2i)(–4 + 6i) \]
\[=2(–4 + 6i) 2i(4+6i)\]

SheldonEinstein
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\[\large{(22i)(4+6i)}\]
\[\large{2(4) 2i(4) 2(6i) 2i(6i)}\]
\[\large{8 + 8 i  12 i 12i^2}\]

surdawi
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>–4 – 4i

SheldonEinstein
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Now since : \[\large{ i = \sqrt{1} }\]
Therefore, \[\large{i^2 = (\sqrt{1})^2 = 1 }\]

SheldonEinstein
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Solving this we get : 8  4i  12 = 44i

SheldonEinstein
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Therefore the answer is : >44i

blessbails
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thank you all so much, i've got it now! @SheldonEinstein @surdawi

SheldonEinstein
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Nice!

UnkleRhaukus
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mistake !

SheldonEinstein
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20 4i

SheldonEinstein
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Sorry it will be : \[\large{8 + 12  4i = 20 4i}\]

SheldonEinstein
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I am REALLY VERY VERY SORRY for that... Thanks @UnkleRhaukus for pointing it out.

UnkleRhaukus
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\[12i^2=+12\]

SheldonEinstein
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Yeah