How do you find the absolute value of #sqrt11+isqrt5#?

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Answer 1 · 2016-08-21T22:12:04

#|sqrt11+isqrt5|=4#

The absolute value of a complex number #a+bi# is written as #|a+bi|# and its value is #sqrt(a^2+b^2)#.

Hence, absolute value of #sqrt11+isqrt5# is

#|sqrt11+isqrt5|#

= #sqrt((sqrt11)^2+(sqrt5)^2)#

= #sqrt(11+5)#

= #sqrt16#

= #4#.