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

Aug 16, 2016

$\left\mid 1 - 4 i \right\mid = \sqrt{17}$

#### Explanation:

For any Complex number $z = x + y i$, $\left\mid z \right\mid$ is the (Euclidean) distance of $z$ from $0$.

Using Pythagoras, the distance is:

$\left\mid x + y i \right\mid = \sqrt{{x}^{2} + {y}^{2}}$

So in our example, we find:

$\left\mid 1 - 4 i \right\mid = \sqrt{{1}^{2} + {\left(- 4\right)}^{2}} = \sqrt{1 + 16} = \sqrt{17}$

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

$\left\mid z \right\mid = \sqrt{z \overline{z}}$

since we have:

$\sqrt{\left(x + y i\right) \overline{\left(x + y i\right)}} = \sqrt{\left(x + y i\right) \left(x - y i\right)} = \sqrt{{x}^{2} + {y}^{2}}$

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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+...))))