How do you write the complex conjugate of the complex number # sqrt(-15)#?

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Answer 1 · 2016-08-01T00:11:17

#bar(sqrt(-15))=-sqrt(15)i#

Given a complex number #a+bi#, the complex conjugate of that number, denoted #bar(a+bi)#, is given by #bar(a+bi) = a-bi#

In our case, then, we have

#bar(sqrt(-15)) = bar(sqrt(15)i)#

#=bar(0+sqrt(15)i)#

#=0-sqrt(15)i#

#=-sqrt(15)i#

Answer 2 · 2016-08-01T00:12:29

#-sqrt(15)i#

It's important to remember that #i^2=-1#

This means we can rewrite

#sqrt(-15) " as " sqrt(15i^2) = sqrt(15)i#

For complex number #z=x+yi# the complex conjugate #bar(z)# is defined as #bar(z) = x-yi#

We have #z = 0 + sqrt(15)i implies bar(z) = 0 - sqrt(15)i#