Algebra with Complex Numbers & Imaginary Unit

[aaaa202]

When i is defined by an equation which has 2 solutions, how does it make sense to do algebra with complex numbers?

 

[1MileCrash]

How is i defined by an equation which has 2 solutions?

Oh, I understand what you mean. I don't think it matters, because of the definition, everything true for i will be true for -i too. It becomes purely a notational issue, we have two solutions, and they are additive inverses, just pick a symbol for one and put a "-" in front of the other to show that it is the additive inverse. It doesn't matter.

 

[aaaa202]

I feel kind of stupid, but I don't understand this argument that if everything is true for i it will be for -i too. Could you give some examples? I feel it's kind of the same as choosing a right or lefthanded coordinate system but then again I have not understood that either.

 

[HallsofIvy]

When i is defined by an equation which has 2 solutions, how does it make sense to do algebra with complex numbers?

It doesn't! We don't define "i" by "$i^2= -1$"- although elementary treatments may introduce it that way.

Rather, we define the complex numbers as pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (a, b)(c, d)= (ac- bd, ad+ bc). That way (a, 0)+ (c, 0)= (a+ c, 0) and (a, 0)(c, 0)= (ac, 0) sp we can associate the "complex number" (a, 0) with the real number, a. Also, if we define i= (0, 1) then we have $i^2= (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0)$ so that "$i^2= -1$". We can then say (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1)= a(1)+ b(i)= a+ bi.

The point is that the way a mathematical concept is introduced to the student is not necessarily the way it is formally defined by mathematicians.

 

[aaaa202]

Okay so maybe you can help me understand where it goes wrong in the following (which motivated this thread for a start):

1. 1/i = sqrt(1)/sqrt(-1)
2. 1/i * i/i = sqrt(1/-1)
3. i / -1 = sqrt(-1)
4. -i = i

What is the problem by simply defining i=sqrt(-1) and where does it go wrong in the above when doing so?

 

[Mentallic]

Okay so maybe you can help me understand where it goes wrong in the following (which motivated this thread for a start):

1. 1/i = sqrt(1)/sqrt(-1)
2. 1/i * i/i = sqrt(1/-1)
3. i / -1 = sqrt(-1)
4. -i = i

What is the problem by simply defining i=sqrt(-1) and where does it go wrong in the above when doing so?

The problem is going from line 1 to 2,

$$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$$

Is a rule that applies only to positive a and b values.

 

[1MileCrash]

I feel kind of stupid, but I don't understand this argument that if everything is true for i it will be for -i too. Could you give some examples? I feel it's kind of the same as choosing a right or lefthanded coordinate system but then again I have not understood that either.

OK, I think it will become obvious to you.

3i + i = 4i
3(-i) + (-i) = 4(-i)

For example, or even
e^(-i(pi)) = -1

Where i/-i is only referenced once.

If we make the 'basis' for the 1 dimensional vector space -i instead of i, the change is only superficial.

Essentially, I believe that -i and i may be regarded as equals (not that they are equal to one another, but that they are on the same footing.)

It doesn't matter which one is which, as long as both exist. I don't believe you can say ANYTHING that differentiates I from -i... but I may be wrong.

 

[aaaa202]

It doesn't! We don't define "i" by "$i^2= -1$"- although elementary treatments may introduce it that way.

Rather, we define the complex numbers as pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (a, b)(c, d)= (ac- bd, ad+ bc). That way (a, 0)+ (c, 0)= (a+ c, 0) and (a, 0)(c, 0)= (ac, 0) sp we can associate the "complex number" (a, 0) with the real number, a. Also, if we define i= (0, 1) then we have $i^2= (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0)$ so that "$i^2= -1$". We can then say (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1)= a(1)+ b(i)= a+ bi.

The point is that the way a mathematical concept is introduced to the student is not necessarily the way it is formally defined by mathematicians.

I see. But what exactly is the difference then between defining i=(0,1) and i=√(-1). Is the only problem that the last one could make you mistakingly think that you can apply usual rules for square roots? And noting that this is not the case, are the 2 definitions the equivalent?

 

[1MileCrash]

It doesn't! We don't define "i" by "$i^2= -1$"- although elementary treatments may introduce it that way.

Rather, we define the complex numbers as pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (a, b)(c, d)= (ac- bd, ad+ bc). That way (a, 0)+ (c, 0)= (a+ c, 0) and (a, 0)(c, 0)= (ac, 0) sp we can associate the "complex number" (a, 0) with the real number, a. Also, if we define i= (0, 1) then we have $i^2= (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0)$ so that "$i^2= -1$". We can then say (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1)= a(1)+ b(i)= a+ bi.

The point is that the way a mathematical concept is introduced to the student is not necessarily the way it is formally defined by mathematicians.

All you've done is arbitrary chosen (0,1) to be i instead of (0, -1). If (0,1) = i, then (0,-1) = -i, and then we're back where we started: -what's the difference?-

 

[lostcauses10x]

aaaa202
try this page.
http://mathforum.org/library/drmath/view/53809.html

 

[rbj]

All you've done is arbitrary chosen (0,1) to be i instead of (0, -1). If (0,1) = i, then (0,-1) = -i, and then we're back where we started: -what's the difference?-

i'm in agreement, Hall, with 1Mile. originally the concept of "imaginary" numbers and then "complex" numbers did come from, i believe, the solution to quadratics (or higher-order polynomials) set to 0 when there were no "real" numbers (the set of rational and irrational numbers that are ordered on the "real" number line) that satisfy those equations.

saying that

$$(a, b) + (c, d) = (a+c, b+d)$$

makes sense just from a normed, linear space POV. but when you say that

$$(a, b) \times (c, d) = (ac-bd, ad+bc)$$

as a definition, we can just as well say that is motivated by or a result of the definition:

$$i^2 = -1$$ .

i might recommend the OP to take a look at the wikipedia article on it, especially the 2nd section. i think that explains it pretty well without just requiring the OP to accept $(a, b) \times (c, d) = (ac-bd, ad+bc)$ as a definition for how complex numbers multiply.

 

[pwsnafu]

originally the concept of "imaginary" numbers and then "complex" numbers did come from, i believe, the solution to quadratics (or higher-order polynomials) set to 0 when there were no "real" numbers (the set of rational and irrational numbers that are ordered on the "real" number line) that satisfy those equations.

This is not true. It comes from the the Tartagalia-Cardano formula, i.e. the cubic formula. As you know cubics always have a real solution, however from time to time the formula generates solutions with square roots of negative numbers. For example $x^3 - 15x = 4$ produces
$x = \sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}}$
but this equals 4.

Another example is in 1702, when Leibniz demonstrated to Hyugens that
$\sqrt6 = \sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}$
who exclaimed "defies all human understanding".

All of this predates the notion of complex numbers.

 

[rbj]

there's a reason i parenthetically inserted "or higher-order polynomials".

but, from a pedagogical POV, i don't understand why it's necessary to go above quadratics to introduce someone conceptually to imaginary and complex numbers. you don't need to go to cubic or quartic to see the need for imaginary and complex numbers.

 

[pwsnafu]

there's a reason i parenthetically inserted "or higher-order polynomials".

You wrote "set to 0 when there were no "real" numbers (the set of rational and irrational numbers that are ordered on the "real" number line) that satisfy those equations." That part is wrong. It's the real solutions that caused the discovery. Back then $x^2 + 1 = 0$ simply had no solutions.

but, from a pedagogical POV, i don't understand why it's necessary to go above quadratics to introduce someone conceptually to imaginary and complex numbers. you don't need to go to cubic or quartic to see the need for imaginary and complex numbers.

But you weren't making a point of pedagogy. You were making a claim about the history of mathematics.

 

[rbj]

no. you cannot represent me nor what i was saying. you cannot even presume to.

it was purely a point of pedagogy from the beginning. as best as i can tell, you were making an historical point. but i was covering my arse anyway.

 

[pwsnafu]

What? You wrote

originally the concept of "imaginary" numbers and then "complex" numbers did come from...

Explain please how that is not about history. You explicitly used the phrase "originally...did come from". How am I supposed to parse that as a pedagogical point?

Look, if you are claiming that teaching complex numbers is easier through quadratics, then I agree with you. But that was not what you wrote in the first paragraph.

 

[lostcauses10x]

Folks is this arguing helping?

 

[pwsnafu]

Folks is this arguing helping?

Yeah. I'm done. Moving on.

About 1Mile's point about i and -i being equals.
As Halls pointed out complex numbers are formally defined as pairs, with $i = (0,1)$. This space of numbers is of course $\mathbb{C}$.

Let's create a new number system. I'll define $j = (0,-1)$ and call this space $\mathbb{D}$ which is composed of numbers $\alpha + \beta j$.

What 1Mile is saying that if I make a statement (theorem) that is true on $\mathbb{C}$, there is an equivalent statement that is true on $\mathbb{D}$. And we can use conjugation to write it down.

 

[rbj]

About 1Mile's point about i and -i being equals.

they are qualitatively equivalent (both have equal claim to squaring to -1), but not quantitatively, since they are not zero and are negatives of each other.

As Halls pointed out complex numbers are formally defined as pairs, with $i = (0,1)$. This space of numbers is of course $\mathbb{C}$.

$\mathbb{C}$ is not $\mathbb{R}^2$. Halls was complete when he stated the rules of addition and multiplication of these pairs. my point is that the "formally defined as pairs" is not the original meaning (in case you're wondering, i mean the original pedagogical meaning), but can be made equivalent as long as you define the rules of addition (trivial, about the only way you can) and multiplication (less trivial, there are other ways to define multiplication of 2-vectors which are not compatible with complex numbers). i really don't see any point to it or advantage over the common pedagogy (geez, i have to be careful to remind snafu, lest the words be misconstrued).

 

[R136a1]

they are qualitatively equivalent (both have equal claim to squaring to -1), but not quantitatively, since they are not zero and are negatives of each other.



$\mathbb{C}$ is not $\mathbb{R}^2$. Halls was complete when he stated the rules of addition and multiplication of these pairs. my point is that the "formally defined as pairs" is not the original meaning (in case you're wondering, i mean the original pedagogical meaning), but can be made equivalent as long as you define the rules of addition (trivial, about the only way you can) and multiplication (less trivial, there are other ways to define multiplication of 2-vectors which are not compatible with complex numbers). i really don't see any point to it or advantage over the common pedagogy (geez, i have to be careful to remind snafu, lest the words be misconstrued).

Getting all argumentative is getting us nowhere, rbj. I think all of pwsnafu's replies have been very accurate.

 

[rbj]

Getting all argumentative is getting us nowhere, rbj. I think all of pwsnafu's replies have been very accurate.

you have every right to think that.

(i'm here for information. $\mathbb{C}$ is not the same as $\mathbb{R}^2$. call it argumentative or not, but if anyone "simplifies" the concept of complex numbers to reducing the concept to only "pairs" of real numbers, they need to read that quote from Einstein: "Things should be as simple as possible, but no simpler.")

 

[lostcauses10x]

The set of complex numbers is not reals squared. True.
And no one has said this.
The statement is ordered pairs of real parts.

Yet by leaving out i it can confuse some, and is easier for others. The questions is "is it equivalent with the definitions for operation defined? I have no doubt of this, yet...

Yet the statement

Rather, we define the complex numbers as pairs of real numbers

does define the rules of the complex system, not the number i.

i of course is undefined in the reals, yet is a constant such as the real part of "a+bi" or bi is such that a defines how many "i"s are added together.

Of course this can confuse folks very easy, and part of the whys "i" were and are avoided by a lot.

 

[rbj]

The set of complex numbers is not reals squared. True.

but the sets are one-to-one with each other. (what did they call that? "entire"? can't remember.)

And no one has said this.

well, don't know if i agree with that. i quoted someone making, without the reference to "$\mathbb{R}^2$", an equivalent statement.

The statement is ordered pairs of real parts.

complex numbers are more than just ordered pairs of real parts. $\mathbb{R}^2$ are also ordered pairs of real parts. while there is a simple and complete one-to-one mapping of $\mathbb{R}^2$ to and from $\mathbb{C}$, it is tempting for some to think that they are the same, but $\mathbb{C}$ requires an operation or rule that $\mathbb{R}^2$ does not. and it makes it different.

Yet by leaving out i it can confuse some, and is easier for others.

and that is solely what my disagreement with Halls is about. it's about pedagogy, and i see little advantage, conceptually, to the pedagogy that leaves out $i^2=-1$.

 

[lostcauses10x]

First to have the ordered pairs, takes two sets of reals; not just one: with the rules as stated in this thread. The intersection of coarse is at (0,0) with one set vertical to the other.
End result is that "i" is equivalent to (0,1)
Which is why I referred to the link I posted.

The problem with boards and such talks is that to state all involved is not practical. Just to get into a talk about a term or terms can lead a topic way off.

And I learned the old way myself. I have always had a bit of trouble when they say (a+bi) that b is a real.
I have had to adapt to all of this myself.

 

[jbriggs444]

but the sets are one-to-one with each other. (what did they call that? "entire"? can't remember.)

bijection? There are several obvious bijections between the two entities, as long as they are viewed as simple sets.

isomorphism? If you tack on the appropriate addition and multiplication operations, there are two obvious isomorphisms between the two enties when viewed as fields.

automorphism? There is one non-trivial automorphism between $\mathbb{C}$ and itself. The mapping is complex conjugation (invert the imaginary part).

 

[lostcauses10x]

Back to the original question.

When i is defined by an equation which has 2 solutions, how does it make sense to do algebra with complex numbers?

Well how does it make sense to use the reals when the square root of 1 has two answers?
(-1) squared= ?, and (1) squared equals what?
Yet both -1 and 1 are two separate numbers.

Oh and the thinking -i is the same as i is also a mistake.
Yes squared they become the same answer, yet -i times i= 1, not -1.

 

[R136a1]

bijection? There are several obvious bijections between the two entities, as long as they are viewed as simple sets.

isomorphism? If you tack on the appropriate addition and multiplication operations, there are two obvious isomorphisms between the two enties when viewed as fields.

automorphism? There is one non-trivial automorphism between $\mathbb{C}$ and itself. The mapping is complex conjugation (invert the imaginary part).

It's a good practice to say isomorphism and automorphism between which specific structures? Fields? Groups? Rings? Topological spaces?

As fields, there are uncountably many automorphisms of $\mathbb{C}$, but only $2$ continuous ones.

 

[rbj]

bijection? There are several obvious bijections between the two entities, as long as they are viewed as simple sets.

isomorphism? If you tack on the appropriate addition and multiplication operations, there are two obvious isomorphisms between the two enties when viewed as fields.

automorphism? There is one non-trivial automorphism between $\mathbb{C}$ and itself. The mapping is complex conjugation (invert the imaginary part).

yah, it's pretty much bijection. for some reason, i don't remember that term for what i would commonly call "one-to-one" or "invertible". i just wanted to include the property in the mapping that in either $\mathbb{C}$ or $\mathbb{R}^2$, there were no orphaned points. "entire" appears to mean something like analytical at every value in $\mathbb{C}$. it's been 35 years, and there's a lot i don't remember.

but regarding the OP's original question, i think the most direct answer is that, while there are two different values of $i$ that have equal claim to satisfying $i^{\ 2} = -1$, it doesn't matter which value you pick as the "positive" $i$. if every lit and textbook were changed and $-i$ and $+i$ were swapped in the language, every theorem would continue to be just as valid. it's sort of like the (mathematical) equivalence to twist; either clockwise or counter-clockwise can be defined as "positive twist" as a convention. or swap left-hand rule with right-hand rule (say, for the cross product) everywhere.

 

[rbj]

Back to the original question.


Well how does it make sense to use the reals when the square root of 1 has two answers?
(-1) squared= ?, and (1) squared equals what?
Yet both -1 and 1 are two separate numbers.

they are qualitatively different. only one of those two numbers are the multiplicative identity.

Oh and the thinking -i is the same as i is also a mistake.
Yes squared they become the same answer, yet -i times i= 1, not -1.

it is no mistake to think of $-i$ and $i$ as qualitatively the same. every property $-i$ has, $+i$ also has.

 

[rbj]

And I learned the old way myself. I have always had a bit of trouble when they say (a+bi) that b is a real.
I have had to adapt to all of this myself.

so, lost, would you have trouble with the meaning of $a+bi$ for $b$ a real and positive integer? i don't know why you would have to adapt very much for that.

 

[lostcauses10x]

"b a real and positive integer"

b a real integer yes.

Not a problem in that statement. Yet it can be confusing due to it is a part of i. Simply put if b =1, and A= 1 then b still is not equal to b. a is only equal to b at (0,0)
Yet the complex number is two sides, which is why I see the process of teaching in the ordered pair coordinates to be of great use.
Since the simplicity of a+bi can and is for every number of the complex, it can be taken to a+b, and of course to (a,b) easily, as long as (a,b) is defined as being of the complex system, with both a and b being numbers of the reals set, of course stating that a is not equal to b except at (0,0)

The complex number is two dimensions, even though I also shorthand to just i or reals at times.
Not a problem to adapt.
I even see a lot of uses for this.

 

[Mark44]

Back to the original question.


Well how does it make sense to use the reals when the square root of 1 has two answers?

We get this question a lot here at PF. The expression $\sqrt{1}$ has one answer: 1. I will grant you that 1 has two square roots, a positive one and a negative one, but the symbol $\sqrt{1}$ represents the principal, or positive, square root of 1.

(-1) squared= ?, and (1) squared equals what?
Yet both -1 and 1 are two separate numbers.

Oh and the thinking -i is the same as i is also a mistake.
Yes squared they become the same answer, yet -i times i= 1, not -1.

 

[Mark44]

First to have the ordered pairs, takes two sets of reals; not just one: with the rules as stated in this thread. The intersection of coarse is at (0,0) with one set vertical to the other.
End result is that "i" is equivalent to (0,1)
Which is why I referred to the link I posted.

The problem with boards and such talks is that to state all involved is not practical. Just to get into a talk about a term or terms can lead a topic way off.

And I learned the old way myself. I have always had a bit of trouble when they say (a+bi) that b is a real.

I don't understand your difficulty. Complex numbers work in a way very similar to two-dimensional vectors. If you have two vectors that don't point in the same direction or in opposite direction, any vector in the plane can be written as a linear combination of your two vectors (that is, as the sum of scalar multiples of the two vectors). In the case of complex numbers, the vectors are "1", a vector that points to the right, and "i", a vector that points straight up. The complex number a + bi is a*1 + b*i, where a and b are real numbers.

I have had to adapt to all of this myself.

 

[rbj]

"b a real and positive integer"

b a real integer yes.

Not a problem in that statement.

so then we extend the concept to negative integers (what that would mean is $-bi + bi = 0$) then extend it to rational $b$, and then hand wave to every real $b$.

Yet it can be confusing due to it is a part of i. Simply put if b =1, and A= 1 then b still is not equal to b.

makes zero sense to me.

a is only equal to b at (0,0)

i guess. but this doesn't really say anything.

Yet the complex number is two sides, which is why I see the process of teaching in the ordered pair coordinates to be of great use.

my objection is to teach complex numbers (which i have done in the context of electrical engineering classes) solely as ordered pairs.

Since the simplicity of a+bi can and is for every number of the complex, it can be taken to a+b,

how? $i$ is not 1.

and of course to (a,b) easily, as long as (a,b) is defined as being of the complex system, with both a and b being numbers of the reals set, of course stating that a is not equal to b except at (0,0)

this almost sounds like you're making sense, but you're not. at least not to me.

The complex number is two dimensions, even though I also shorthand to just i or reals at times.
Not a problem to adapt.
I even see a lot of uses for this.

the problem with teaching complex numbers without $i$ (and without $i^{\ 2} = -1$) is that you must present as an axiom that

$$(a,b) \times (x,y) \triangleq (ax-by, bx+ay)$$

besides the addition axiom

$$(a,b) + (x,y) \triangleq (a+x, b+y)$$

the latter which is pedagogically trivial. but the former is not. the first thing students will ask is "why the he11 define multiplication like that?" how do you answer that? you might say, "that's the only way we can make the distributive property work" and then proceed to prove it, but it's far more complicated than starting out with $i$ such that $i^{\ 2} = -1$.

 

[1MileCrash]

Back to the original question.


Well how does it make sense to use the reals when the square root of 1 has two answers?
(-1) squared= ?, and (1) squared equals what?
Yet both -1 and 1 are two separate numbers.

1 is not defined to be the square root of 1.

Suppose we inverted the negative and positive imaginary numbers, and left all arithmetic the same. Is everything that was true before, still true? I say yes.

Now suppose we inverted the negative and positive real numbers, and left all arithmetic the same. Then statements do not remain true. For example, if this behaved like switching the imaginary signs, the true statement 3^(1) = 3 would be changed into (-3)^(-1) = (-3), but that is false.

But note that for switching sign of the imaginaries:
3^(i) = A + Bi for the appropriate reals A, B
and
3(-i) = A - Bi = A + B(-i)

So the statement 3^(i) = A + Bi remains true when the sign of i is changed.

Oh and the thinking -i is the same as i is also a mistake.
Yes squared they become the same answer, yet -i times i= 1, not -1.

I don't see your point.

(-i)(i) = 1 says equal things about (-i) and (i). This doesn't show any difference between the two (by "difference," I mean some statement that is true for one and false for the other). The only difference is "call one negative, call one positive." That's why in my example above, changing which ones we call negative and positive has no effect on the truth value of any statement.

The only difference between i and -i is "relative to each other." They are truly two sides of a coin.