How do you find $∣ ∣ - 4 \sqrt{3} - 4 i ∣ ∣$ $abs( -4sqrt3-4i )$ ?

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Answer 1 · 2016-05-23T14:04:11

$∣ ∣ - 4 \sqrt{3} - 4 i ∣ ∣ = 8$ $|-4sqrt3-4i|=8$

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

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

Hence absolute value $∣ ∣ - 4 \sqrt{3} - 4 i ∣ ∣$ $|-4sqrt3-4i|$ is

$\sqrt{{(4 \sqrt{3})}^{2} + 4^{2}} = \sqrt{48 + 16} = \sqrt{64} = 8$ $sqrt((4sqrt3)^2+4^2)=sqrt(48+16)=sqrt64=8$

How do you find ∣∣−4√3−4i∣∣abs( -4sqrt3-4i )?