What is the irrational conjugate of #1+sqrt8#? complex conjugate of #1 + sqrt(-8)#?

1 Answer
Aug 15, 2016

Answer:

#1-sqrt 8 and 1-sqrt(-8)=1-i sqrt 8#, where i symbolizes #sqrt(-1)#.

Explanation:

The conjugate of the irrational number in the form

#a+bsqrt c#,

where c is positive and a, b and c are rational (including computer

string-approximations to irrational and transcendental numbers) is

a-bsqrt c#'

When c is negative, the number is termed complex and the

conjugate is

#a+ibsqrt(|c|)#, where# i = sqrt(-1)#.

Here, the answer is

#1-sqrt 8 and 1-sqrt(-8)=1-i sqrt 8#, where i symbolizes #sqrt(-1)#