How do you find $abs( -4sqrt3-4i )$ ?

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Answer 1 · 2016-05-23T14:04:11

$|-4sqrt3-4i|=8$

Absolute value of a complex number $a+ib$ is denoted as $|a+ib|$

and is given by $sqrt(a^2+b^2)$

Hence absolute value $|-4sqrt3-4i|$ is

$sqrt((4sqrt3)^2+4^2)=sqrt(48+16)=sqrt64=8$

How do you find abs( -4sqrt3-4i )abs( -4sqrt3-4i )?