MHB Real conjugates of complex numbers?

[samir]

Hi!

We have discussed complex numbers in class and their conjugates. From what I understand only the imaginary unit is conjugated. But I wonder if there are such things as real conjugates of complex numbers?

Given the following points:

$$A=(-2+i)$$
$$B=(2+3i)$$
$$C=(-4-3i)$$
$$D=(-4+i)$$

I would argue that their imaginary conjugates are:

$$\overline{A}=(-2-i)$$
$$\overline{B}=(2-3i)$$
$$\overline{C}=(-4+3i)$$
$$\overline{D}=(4-i)$$

I would argue that their real conjugates are:

$$\overline{A}=(2+i)$$
$$\overline{B}=(-2+3i)$$
$$\overline{C}=(4-3i)$$
$$\overline{D}=(4+i)$$

Are these valid statements? Do we ever seek real conjugates of complex numbers? If not, why not? If a problem description only states that we seek a "conjugate" of a complex number, is it always implied that it is the imaginary conjugate we seek?

I have plotted the imaginary conjugates in the following image. (You will have to "imagine" the real conjugates.)

View attachment 5481

 

[Deveno]

I think I understand what you are trying to say, but no: such a concept is hardly ever used.

One can think of conjugation as reflection about the real axis, and (from symmetry) one might want to investigate the usefulness of reflection about the imaginary axis.

As far as *addition* goes, they appear to be on equal footing (this is in keeping with regarding complex numbers as a two-dimensional vector space over the reals-that is, the Argand plane). But the reflection you are considering isn't as useful as "normal" conjugation for this very important reason:

Complex conjugation respects complex multiplication AS WELL AS addition. That is:

$\overline{zw} = \overline{z}\cdot \overline{w}$.

This is NOT the case for the "other reflection".

For example, let's multiply $2 + 3i$ and $1 - i$:

$(2 + 3i)(1 - i) = 2 - 2i + 3i -3i^2 = 2 - i - (-3) = 5 + i$.

Reflecting across the imaginary axis, we have $-5 + i$.

If we reflect both complex numbers first, we have: $-2 + 3i$ and $-1 - i$. Now we multiply and get:

$(-2 + 3i)(-1 - i) = (-2)(-1) + (-2)(-i) + 3i(-1) - 3i^2 = 2 + 2i - 3i - (-2) = 5 - i$.

As you can see the two results are not equal.

It turns out that the map $z \mapsto \overline{z}$ (or $a + ib \mapsto a - ib$, if you prefer) preserves the field structure of the complex numbers, and $a + ib \mapsto -a + ib$ does not. Put another way, in $\Bbb C$, the complex numbers $i$ and $-i = i^3 = \dfrac{1}{i}$ are interchangeable (if we do so "globally"), but $1$ and $-1$ have DIFFERENT algebraic properties ($1$ is the multiplicative identity, $-1$ is not), and switching them "ruins multiplication".