How do you find #abs( -6-8i )#?

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Answer 1 · 2016-04-10T07:51:44

#abs(-6-8i) = 10#

The modulus of a Complex number is essentially its distance from the number #0# in the Complex plane.

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)# is:

#abs(z) = sqrt(z bar(z))#

where #bar(z)# is the complex conjugate of #z#.

Notice that:

#(a+bi)bar((a+bi)) = (a+bi)(a-bi) = a^2-b^2i^2 = a^2+b^2#