What is complex conjugate of i?

2 Answers
Jul 3, 2018

-i

Explanation:

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

Jul 3, 2018

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).