How do you find $abs( 9+i )$ ?

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Answer 1 · 2016-03-23T00:18:32

$abs(9+i) = sqrt(82) ~~ 9.055385$

$abs(a+bi)$ is essentially the distance between $a+bi$ and $0$ in the Complex plane.

From Pythagoras, we get the distance formula and hence find:

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

Another way of expressing this is that $abs(z) = sqrt(z bar(z))$ (where $bar(z)$ means the Complex conjugate of $z$ ).

To see this, notice that:

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

In our example,

$abs(9+i) = sqrt(9^2+1^2) = sqrt(81+1) = sqrt(82) ~~ 9.055385$

How do you find abs( 9+i )abs( 9+i )?