How do you find #abs( -6-8i )#?
1 Answer
Apr 10, 2016
Explanation:
The modulus of a Complex number is essentially its distance from the number
By Pythagoras theorem that means that:
#abs(a+bi) = sqrt(a^2+b^2)#
In our example,
#abs(-6-8i) = sqrt((-6)^2+(-8)^2) = sqrt(36+64) = sqrt(100) = 10#
Another formula for
#abs(z) = sqrt(z bar(z))#
where
Notice that:
#(a+bi)bar((a+bi)) = (a+bi)(a-bi) = a^2-b^2i^2 = a^2+b^2#