How do you find $abs( 1-4i )$ ?

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Answer 1 · 2016-08-16T19:55:22

$abs(1-4i) = sqrt(17)$

For any Complex number $z = x+yi$ , $abs(z)$ is the (Euclidean) distance of $z$ from $0$ .

Using Pythagoras, the distance is:

$abs(x+yi) = sqrt(x^2+y^2)$

So in our example, we find:

$abs(1-4i) = sqrt(1^2+(-4)^2) = sqrt(1+16) = sqrt(17)$

Another way to express the norm of $z$ is:

$abs(z) = sqrt(zbar(z))$

since we have:

$sqrt((x+yi)bar((x+yi))) = sqrt((x+yi)(x-yi)) = sqrt(x^2+y^2)$

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Random footnote

$17$ happens to be my favourite number:

It is the number of possible symmetries of wallpaper patterns (i.e. biperiodic tessellations of the plane).
It is one of the few Fermat primes, being a prime number of the form $2^(2^n) + 1$ . Pierre de Fermat conjectured that all such numbers are prime, but it fails for $n = 5$ and all known greater values.
It is the smallest positive integer (apart from $1$ ) that is expressible as the sum of a square and a cube in two distinct ways:

$17 = 2^3+3^2 = 4^2+1^3$
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It has a nice continued fraction for its square root:

$sqrt(17) = [4;bar(8)] = 4+1/(8+1/(8+1/(8+1/(8+...))))$

How do you find abs( 1-4i )abs( 1-4i )?