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)#

#color(white)()#
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#
#color(white)()#
It has a nice continued fraction for its square root:

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