What is the complex conjugate of #sqrt(-7)#?

1 Answer
Dec 8, 2015

Answer:

#-isqrt7#

Explanation:

To find a complex conjugate, simply change the sign of the imaginary part (the part with the #i#). This means that it either goes from positive to negative or from negative to positive.

As a general rule, the complex conjugate of #a+bi# is #a-bi#.

Your case is seemingly odd, and may not seem to fit the pattern.

However, #sqrt(-7)=isqrt7#.

As such, this can be written in the #a+bi# form of a complex number as #0+isqrt7#.

Thus, the complex conjugate of #0+isqrt7# is #0-isqrt7#, which equals #-isqrt7#.