How do you find #abs( sqrt11 + i sqrt(5))#?

1 Answer
sente
May 30, 2016

Use the definition of a modulus #|a+bi| = sqrt(a^2+b^2)# to find that

#|sqrt(11)+sqrt(5)i|=4#

Explanation:

Given a complex number #a+bi#, the modulus of that number, denoted #|a+bi|#, is given by

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

This is analogous to the absolute value of a real number, giving the distance of the complex number from the origin on the complex plane, rather than the distance from #0# on the number line.

For our given value, then, we have

#|sqrt(11)+sqrt(5)i| = sqrt((sqrt(11))^2+(sqrt(5))^2)#

#=sqrt(11+5)#

#=sqrt(16)#

#=4#

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