Definition of Complex Conjugate

[Jest3r]

Hey all, I was just curious:

Why is the conjugate of a complex number (a + bi) defined as (a - bi)? If we instead change the sign of the real part (-a + bi), we still get a real number when we multiply the two. Is there a particular significance to the current definition?

Thanks a lot for your time!

 

[Doc Al]

For one thing, you want the product of a complex number and its conjugate to be positive.

 

[flatmaster]

a is an arbitrary real number anyway, so -a is redundant.

 

[m00npirate]

b is an arbitrary real number as well, he's asking why you reverse the sign of b and not a, for which i do not have an answer. my best guess is that its just more useful to define it that way.

 

[HallsofIvy]

Doc Al gave the correct response. With \overline{z}= a- bi, the product z*\overline{z}= a^2+ b^2> 0 so we can define |z|= \sqrt{z\overline{z}}=\sqrt{a^2+ b^2}, the distance from z, as a point in the complex plane, to 0.

If we defined \overline{z}= -a+ bi, then z\overline{z} would be equal to -a^2- b^2 and would have to use |z|= \sqrt{-z\overline{z}}.

Of course, this whole thread is a question, not a "learning material" so I am moving it to "General Math".

 

[Hurkyl]

The formula for the absolute value of a complex number is hardly a motivating example.


Somewhere at the start of complex analysis, we said "Let i be a square root of -1", and we built everything on top of that. However, we could have picked the other square root of -1, and worked with that one instead. Complex conjugation swaps back and forth between the two possibilities.

 

[deferro]

I use the mnemonic that the 'circle' has a positive and a negative surface, which meet at the x axis, so y = +/- i

You just use an ordinary circle and say the ordinal is imaginary (fold a circle of paper in half, now it has two imaginary surfaces hidden 'inside'.

 

[adriank]

Hurkyl has a good point. Let me try to describe further.

The important properties of complex conjugation are that (1) it is an automorphism of C, and (2) it fixes R. (An automorphism of C is a one-to-one and onto function f: C → C such f(z + w) = f(z) + f(w) and f(zw) = f(z)f(w) for all z and w in C; f fixes R if f(z) = z for all z in R.) Moreover, it is the unique nontrivial one (that is, it is the only one that is not the identity function). The map a + bi ↦ -a + bi is not an automorphism (if f(a + bi) = -a + bi, then f(i)² = i² = -1, but f(i²) = f(-1) = 1).

Proof: You already know that conjugation is an automorphism of C that fixes R. To prove uniqueness, let f be an automorphism of C that fixes R. Then f(-1) = -1. But f(-1) = f(i²) = f(i)², so f(i)² = -1. Thus, f(i) = i or f(i) = -i. Now for any z in C, write z = a + bi, where a and b are in R. Then f(z) = f(a + bi) = f(a) + f(b)f(i) = a + bf(i). If f(i) = i, then f(z) = a + bi = z, so f is the identity. If f(i) = -i, then f(z) = a - bi, which is the conjugate of z.

 

[CRGreathouse]

Good explanation, Hurkyl.