What is complex conjugate of #i#?

2 Answers
Jul 3, 2018

Answer:

#-i#

Explanation:

Conjugate of any complex number #a+bi# is #a-bi#

Jul 3, 2018

Answer:

It depends...

Explanation:

A conjugate of a number is a number that goes with it in the sense that multiplying the two numbers together yields a simpler kind of number.

For numbers involving radicals, we want to find a multiplier that results in a rational result. For complex numbers we want to find a multiplier that lead to a real result.

If #a+bi# is any complex number (where #a, b in RR#) then a suitable conjugate number is #a-bi#.

This is popularly known as "the" complex conjugate of #a+bi# and we write:

#bar(a+bi) = a-bi#

We find:

#(a+bi)(a-bi) = a^2-b^2i^2 = a^2+b^2 in RR#

Using this convention, we find:

#bar(i) = bar(0+1i) = 0-1i = -i#

However, please note that this is not the only possible choice of conjugate.

For example:

#i * i = -1 in RR#

So we could call #i# self-conjugate.

In fact we could choose any pure imaginary number to use as a conjugate (in the multiplicative sense) for #i#.

Why do I stress this ambiguity?

Consider #sqrt(2)+sqrt(3)#.

What would you say is "the" radical conjugate of #sqrt(2)+sqrt(3)# (i.e. the natural choice of multiplier to give a rational product) ?

We could mechanically choose #sqrt(2)-sqrt(3)# and find:

#(sqrt(2)-sqrt(3))(sqrt(2)+sqrt(3)) = 2-3 = -1 in QQ#

So #sqrt(2)-sqrt(3)# is certainly a conjugate, but somewhat nicer is #sqrt(3)-sqrt(2)# ...

#(sqrt(3)-sqrt(2))(sqrt(2)+sqrt(3)) = 3-2 = 1#

Going back to complex conjugates, the standard complex conjugate #bar(a+bi) = a-bi# is significant for other reasons than being a multiplicative conjugate. For example, if #a+bi# is a zero of a polynomial with real coefficients then #bar(a+bi) = a-bi# is also a zero.

So in general usage, it is conventional to stay with the standard complex conjugate and refer to "complex conjugate pairs" (for example).