What is the conjugate of #sqrt(-20)#?

1 Answer
sente
Nov 21, 2016

#-2sqrt(5)i#

Explanation:

Given a complex number #z=a+bi# (where #a, b in RR# and #i = sqrt(-1)#), the complex conjugate or conjugate of #z#, denoted #bar(z)# or #z^"*"#, is given by #bar(z) = a-bi#.

Given a real number #x>=0#, we have #sqrt(-x) = sqrt(x)i#.

note that #(sqrt(x)i)^2 = (sqrt(x))^2*i^2 = x*-1 = -x#

Putting these facts together, we have the conjugate of #sqrt(-20)# as

#bar(sqrt(-20)) = bar(sqrt(20)i)#

#=bar(0+sqrt(20)i)#

#=0-sqrt(20)i#

#=-sqrt(20)i#

#=-2sqrt(5)i#

Related questions
See all questions in Division of Complex Numbers
Impact of this question
6640 views around the world
You can reuse this answer
Creative Commons License