How do you find $∣ ∣ 1 - i \sqrt{3} ∣ ∣$ $abs( 1- i sqrt3)$ ?

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Answer 1 · 2016-05-01T05:42:24

$∣ ∣ 1 - i \sqrt{3} ∣ ∣ = 2$ $|1-isqrt3|=2$

$| a + i b | = \sqrt{a^{2} + b^{2}}$ $|a+ib|=sqrt(a^2+b^2)$

Hence $∣ ∣ 1 - i \sqrt{3} ∣ ∣ = \sqrt{1^{2} + {(- \sqrt{3})}^{2}} = \sqrt{1 + 3} = \sqrt{4} = 2$ $|1-isqrt3|=sqrt(1^2+(-sqrt3)^2)=sqrt(1+3)=sqrt4=2$

How do you find ∣∣1−i√3∣∣abs( 1- i sqrt3)?