How do you find #abs( 14i )#?
1 Answer
Explanation:
For any Complex number
Using Pythagoras, the distance is:
#abs(x+yi) = sqrt(x^2+y^2)#
So in our example, we find:
#abs(14i) = sqrt(1^2+(4)^2) = sqrt(1+16) = sqrt(17)#
Another way to express the norm of
#abs(z) = sqrt(zbar(z))#
since we have:
#sqrt((x+yi)bar((x+yi))) = sqrt((x+yi)(xyi)) = sqrt(x^2+y^2)#
Random footnote

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+...))))#