How do you find #abs( 9+i )#?

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Answer 1 · 2016-03-23T00:18:32

#abs(9+i) = sqrt(82) ~~ 9.055385#

#abs(a+bi)# is essentially the distance between #a+bi# and #0# in the Complex plane.

From Pythagoras, we get the distance formula and hence find:

#abs(a+bi) = sqrt(a^2+b^2)#

Another way of expressing this is that #abs(z) = sqrt(z bar(z))# (where #bar(z)# means the Complex conjugate of #z#).

To see this, notice that:

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

In our example,

#abs(9+i) = sqrt(9^2+1^2) = sqrt(81+1) = sqrt(82) ~~ 9.055385#