How do you find the absolute value of $-8i$ ?

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Answer 1 · 2017-01-31T10:24:18

$8$

Given a complex number like $z=x+iy$ then $abs z$ is defined as

$abs z = sqrt(z cdot \bar z$ where $bar z = x-iy$ is the $z$ conjugate.

Then $absz=sqrt(x^2+y^2)$ . In our case

$abs(-i8) = sqrt((-i8)(i8))$ = 8

$abs z ge 0$ by definition.

How do you find the absolute value of -8i-8i?