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Question #f4b39

1 Answer

May 5, 2016

$|z|=sqrt29$

Explanation:

$"let be ; z=u+v"$

$z=(-5+10)i+(1+1)j$

$z=5i+2j$

$|z|=sqrt(5^2+2^2)$

$|z|=sqrt(25+4)$

$|z|=sqrt29$

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