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Question #df1ef

1 Answer

Nov 23, 2016

$P(1-3i)=-13-6i$

Explanation:

$P(x)=x^2-5$

$P(1-3i)=(1-3i)^2-5$
$=(1-3sqrt(-1))(1-3sqrt(-1))-5$
$=1-3sqrt(-1)-3sqrt(-1)+9(-1)-5$
$=1-6i-14$
$=-13-6i$

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