Question

What is complex conjugate of `i`?

Answer 1

`-i`

Explanation:

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

Answer 2

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).