Conjugate vs Complex Conjugate

[rstor]

Homework Statement

I would like to confirm my understanding between conjugates and complex conjugates.

Relevant Equations

not applicable

I would like to confirm my understanding between conjugates and complex conjugates. When rationalizing the denominator (when dealing with square roots) it is shown that we can multiply the fraction by its conjugate. I have seen the conjugate being defined as one of the signs of the binomial changing (does not have to be the middle sign).
Example:

With regards to the Complex Conjugate my understanding is that the sign of the imaginary part must change (the middle sign when the complex number is written in standard form).

Is this correct?

 

[Mark44]

Homework Statement: I would like to confirm my understanding between conjugates and complex conjugates.
Relevant Equations: not applicable

I would like to confirm my understanding between conjugates and complex conjugates. When rationalizing the denominator (when dealing with square roots) it is shown that we can multiply the fraction by its conjugate. I have seen the conjugate being defined as one of the signs of the binomial changing (does not have to be the middle sign).
Example:

With regards to the Complex Conjugate my understanding is that the sign of the imaginary part must change (the middle sign when the complex number is written in standard form).

Is this correct?

Pretty much.

The conjugate of $a + b$ is $a - b$. The complex conjugate of $a + ib$ is $a - ib$.
The real numbers a and b might both be positive, both negative, or mixed in sign.
Note that we don't normally call the imaginary unit "iota". It's normally called "i", short for imaginary.

 

[FactChecker]

With regards to the Complex Conjugate my understanding is that the sign of the imaginary part must change (the middle sign when the complex number is written in standard form).

Is this correct?

That certainly is the correct definition of the complex conjugate. And with complex matrices, the conjugate matrix is obtained by taking the complex conjugate of every element. It's my understanding that the idea of the complex conjugate has been mimicked by the conjugate in other algebraic uses, but I have no experience with that.

 

[rstor]

Pretty much.

The conjugate of $a + b$ is $a - b$. The complex conjugate of $a + ib$ is $a - ib$.
The real numbers a and b might both be positive, both negative, or mixed in sign.
Note that we don't normally call the imaginary unit "iota". It's normally called "i", short for imaginary.

What had confused me was that to my understanding you can write the conjugate of a+b as a-b OR -a+b however the complex conjugate of a+bi can only be written as a-bi and
not -a+bi

 

[FactChecker]

What had confused me was that to my understanding you can write the conjugate of a+b as a-b OR -a+b

Apparently, that is a general definition in algebra. I don't know how they apply it, but the basic idea is to change the sign of one of two independent components without changing the other. This can be handy. You can see some uses in complex analysis: if $z=a+ib$, and $\bar{z}=a-ib$ denotes the complex conjugate, then $Real(z) = a = (z+\bar{z})/2$. So the components of $z$ can be represented using simple arithmetic.

however the complex conjugate of a+bi can only be written as a-bi and
not -a+bi

Correct. The complex conjugate has a very special role in that it negates the angle of a complex number from the positive real line. So the argument of $\bar{z}$ is the negative of the argument of $z$. That restricts how the complex conjugate is defined. But then it allows us to say that $z\bar{z} = |z|^2$ is always a non-negative real number. The feature of complex numbers where angles and rotations can be represented and manipulated by multiplication is very profound and fundamental.

 

[Mark44]

my understanding you can write the conjugate of a+b as a-b OR -a+b

You can rewrite a + b as b + a, so its conjugate is b - a. The important point is that you want the product to be the difference of two squares; i.e., $a^2 - b^2$ or in the revised case, $b^2 - a^2$. In this way you can eliminate square roots.
With a complex number a + bi, when you multiply by the complex conjugate, you still get a difference of squares, sort of, in the form of $a^2 - i^2b^2$, or more simply, $a^2 + b^2$, and you now have a real number.

 

[pasmith]

What had confused me was that to my understanding you can write the conjugate of a+b as a-b OR -a+b however the complex conjugate of a+bi can only be written as a-bi and
not -a+bi

What are $a$ and $b$ in your first example? Are they the same kind of object? $a$ and $ib$ in your second example are different types of object: $a$ is a real number, but $ib$ is not.

Given a field $K$, for $\Delta \notin K$ but $\Delta^2 \in K$ we can define the field $K[\Delta] = \{x + y\Delta : (x,y) \in K^2 \}$ with the following operations: $$\begin{split} (x_1 + y_1\Delta) + (x_2 + y_2\Delta) &= (x_1+x_2) + (y_1 + y_2)\Delta, \\ (x_1 + \Delta y_1)(x_2 + y_2\Delta) &= (x_1x_2 + y_1y_2\Delta^2) + (x_1y_2 + x_2y_1)\Delta.\end{split}$$ The complex numbers are a special case of this with $K = \mathbb{R}$ and $\Delta^2 = -1$. Another case which arises frequently in applications is $K = \mathbb{Q}$ and $\Delta^2 > 0$ is a rational number whose square root $\Delta > 0$ is irrational.

We see from the rules above that $$(x + y\Delta)(x - y\Delta) = x^2 - y^2\Delta^2 \in K.$$ Now $$(x + y\Delta)(-x + y\Delta) = y^2\Delta^2 - x^2 \in K,$$ but the map $$\phi : K[\Delta] \to K[\Delta] : x + y\Delta \mapsto x - y\Delta$$ fixes $K$, whereas the map $$\psi : K[\Delta] \to K[\Delta] : x + y\Delta \mapsto -x + y\Delta$$ is $-\phi$, and so fixes only $\{0\}$. This is suffcient fur us to prefer to define $\phi$ as the conjugation map.

Note that if $K$ is ordered then $x^2 - y^2\Delta^2$ is not necessarily positive when $\Delta^2 > 0$, although it can never be zero. Generally we prefer positive denominators and negative numerators, so that might govern which of $$\begin{split} (x + y\Delta)^{-1} &= \frac{x}{x^2 - y^2\Delta^2} - \frac{y}{x^2 - y^2\Delta^2}\Delta \\ &= -\frac{x}{y^2\Delta^2 - x^2} + \frac{y}{y^2 \Delta^2 - x^2}\Delta\end{split}$$ we use in a given case.

 

[FactChecker]

What are $a$ and $b$ in your first example? Are they the same kind of object? $a$ and $ib$ in your second example are different types of object: $a$ is a real number, but $ib$ is not.

That is a very good point. For the conjugate to be beneficial, the two components are NOT things that can be simply added as though they are real numbers. They should be linearly independent vectors like $a+ib$, $x+x^2$, $sin()+cos()$, etc.

 

[mathwonk]

note in pasmith's nice discussion that -phi also fixes, not only 0, but all multiples of delta, i.e. all elements of K[delta] with x=0, but this fixed set does not form a subfield, i.e. is not closed under addition and multiplication. A nice property of phi, that causes its fixed set to be a subfield, is that, unlike -phi, it preserves the operations of addition and multiplication. I.e. the new field K[delta] has a symmetry to its arithmetic, in that delta can be exchanged with -delta, preserving the arithmetic operations. So we get an "action" of the group Z/2Z acting as symmetries on this new field, in such a way as to leave fixed the subfield K.
Anytime we have a group action on anything, the various images of an element under the action may be considered "conjugates" of the original element. It is sort of like a "lazy Susan" in a restaurant, where the items around the outside are all conjugate, since their positions can be interchanged by a rotation, but the dish in the center cannot be, hence is conjugate only to itself, or "fixed".
https://en.wikipedia.org/wiki/Lazy_...rving_turntable_restaurant_in_China,_1987.jpg

An example with such a cyclic structure, is the field of solutions of X^5-1, whose solutions are 1, w, w^2, w^3, w^4, where w = e^i(2π/5) is a "primitive" 5th root of 1. All these roots are powers of any one of them except 1, and the symmetries of the roots in this field look like the lazy Susan with 1 fixed in the middle and the 4 others (all conjugate) arranged in order around the outer circle.

Here is the simplest example I could think of involving just three conjugates: if x is a (real) cube root of 2, and if y and z are the other two (complex) cube roots of two, then just adjoining x to the rational numbers gives us a field whose elements have form a + bx + cx^2, where a,b,c are rational.
In the smallest field containing all cube roots of 2, all three cube roots of 2 are conjugate to each other, so this element a + bx + cx^2 has two other conjugates, namely a + by + cy^2, and a + bz + cz^2.

[If you are like me, you may think there should be something special about the real cube root, but in fact in this field there isn't. I.e. although there is a symmetry of this field exchanging the two complex cube roots, there are also symmetries interchanging the real root with either one of the complex ones. I.e., in terms of the original question, y and z are complex conjugates here, but x and y are also conjugate by a different symmetry, just not complex conjugation, as are also x and z. There is also a cyclic symmetry here permuting x,y,z to say y,z,x. In fact in this field, every permutation of the 3 roots preserves the arithmetic operations.]

Then to "rationalize" the denominator of a fraction with a+bx+cx^2 in the bottom, one could then multiply bottom and top by the product of both the other two conjugates;
i.e. the product (a+bx+cx^2)(a+by+cy^2)(a+bz+cz^2)
is guaranteed to be a rational number, even though x is irrational, and y and z are both "irrational" and complex.
In fact any polynomial expression in x,y,z, and their squares, with rational coefficients, which is "symmetric" in x,y,z, i.e. whose value is unchanged when these letters are permuted, will actually be a rational number. It follows that also the sum of the three conjugates is rational, (in fact x+y+z = x^2 + y^2 + z^2 = 0, so the sum of the conjugates above is 3a. [oops, not a+b+c as I said earlier]), but this is less useful for rationalizing a fraction.

[Recall that the rational coefficients of the polynomial X^3-2, satisfied by the roots x,y,z are ± the values of the "elementary" symmetric functions on these roots, namely x+y+z, xyz, and xy+xz+yz, so in particular x+y+z = 0, and xyz = 2; and all symmetric functions are rational expressions of these elementary ones, so in fact every symmetric function of the roots is rational.]

This "theory of equations" leads to what is now called Galois theory, and I am not an expert, but I have enjoyed teaching it at least once. Here is what looks like a beautiful, and highly readable, set of notes from someone (Tom Leinster, at U of Edinburgh) who actually understands the subject, and can hence explain it clearly: https://www.maths.ed.ac.uk/~tl/gt/gt.pdf

E.g. already in the first few pages he gives a simple way to think of conjugacy: namely two complex numbers x,y are conjugate over the rationals if x satisfies every polynomial with rational coefficients that y satisfies. In our example, the simplest rational polynomial that x satisfies is X^3-2, and so also do y and z satisfy this polynomial. (Probably LaGrange, whose work Galois studied, already understood as much of the theory of equations as we are using here.)

There are many interesting and surprising things about this topic, at least to me. E.g. in the related thread (linked below) on rationalizing a denominator containing a cube root, note that the solution was to multiply the denominator (and numerator) by the square of the cube root. Since this is obviously the easiest answer, one might wonder why my method above is to multiply by the product of the other two cube roots. The surprising result is that these are the same! I.e. the square of the real cube root of 2, equals the product of the two complex cube roots of 2! Although it surprised me, this follows immediately from the fact that the product of all three roots of the polynomial X^3-2 equals (minus) the constant term, i.e. 2.

https://www.physicsforums.com/threa...tor-if-the-denominator-is-a-cube-root.705651/

 

[mathwonk]

some more detail:
let F = Q(t) where t^2 =2. then elements of F have form z = a+bt, for rational a and b. define z’ = a-bt to be the conjugate of z.

then claim if z = a +bt and w = c+dt, then z’w’ = (zw)’, i.e. the conjugate of a product of elements of F equals the product of their conjugates.
proof: zw = ac + (ad+bc)t + bdt^2 = (ac+2bd) + (ad+bc)t,
hence (zw)’ = (ac+2bd) - (ad+bc)t.

next, z’w’ = ac - (ad+bc)t + bdt^2
= (ac + 2bd) - (ad+bc)t. so these are equal.

on the other hand, if we define z* = -a+bt, then w* = -c+dt, then z*w* = (ac+2bd) -(ad+bc)t.

however (zw)* = [(ac+2bd) + (ad+bc)t]*
= -(ac+2bd) + (ad+bc)t. so (zw)* = -z*w*.

hence * does not commute with multiplication, while ' does. hence the preferred definition of conjugation is ‘ and not *.

well after all that, I guess the fact that * sends 1 to -1 is already the issue, i.e. that is not a good indication for a proposed automorphism. as usual I tend not to notice the short way to see something until after doing it the long way.