How do you find $abs( -6-8i )$ ?

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Answer 1 · 2016-04-10T07:51:44

$abs(-6-8i) = 10$

The modulus of a Complex number is essentially its distance from the number $0$ in the Complex plane.

By Pythagoras theorem that means that:

$abs(a+bi) = sqrt(a^2+b^2)$

In our example,

$abs(-6-8i) = sqrt((-6)^2+(-8)^2) = sqrt(36+64) = sqrt(100) = 10$

Another formula for $abs(z)$ is:

$abs(z) = sqrt(z bar(z))$

where $bar(z)$ is the complex conjugate of $z$ .

Notice that:

$(a+bi)bar((a+bi)) = (a+bi)(a-bi) = a^2-b^2i^2 = a^2+b^2$

How do you find abs( -6-8i )abs( -6-8i )?