Complex numbers

[wandelaar]

Why can't I add whitesilk to my ignore list?

 

Edited August 29, 2018 by wandelaar

[OldDog]

Sooo, we are to treat i simply as a variable and defer the notion of imaginary? 

[wandelaar]

The "i" is only a sign (not a variable) used in the complex notation (a + bi) of arrows. However there is an arrow with the complex notation 0 + 1i that will later on play the role of the "imaginary unit". And in shorthand that arrow will often be written as "1i" or even simply as "i".

 

Now there is no real number x such that x*x = -1. But as we will see later on there are complex numbers z (and in our geometric approach arrows z) such that z*z = -1 + 0i . That is the way mathematicians solve this issue. They create a larger number system (namely that of the complex numbers) in which there are solutions z such that z*z = -1 + 0i. And those solutions aren't figments of the imagination but concrete objects (arrows in our case) from the set of complex numbers.

Edited August 29, 2018 by wandelaar

[Marblehead]

Well, I back.  Doesn't mean I am ready for more input but we will see.

 

[wandelaar]

It's up to you. Do you want some more training before moving on to the next phase? Or do you have some questions left?

[Marblehead]

7 minutes ago, wandelaar said:

It's up to you. Do you want some more training before moving on to the next phase? Or do you have some questions left?

Yes, I still have questions but I don't know what they are.  How's that for being insecure?

 

So yes, I can plot a point on a graph -1,3

 

And I can draw an arrow from point 0,0 to -1,3 and define it as arrow -1,3i.

 

But if you ask me what I have done all I can say is that I drew an arrow from 0,0 to -1,3 and called that arrow -1,3i.

 

If you ask me what that means I will respond that you have asked me one too many questions.

 

 

But yes, we can continue if you think I am ready for it.  Remember, I never said this was going to be easy.

 

[Lost in Translation]

2 hours ago, wandelaar said:

there are solutions z such that z*z = -1 + 0i

 

?

 

Please continue your lesson. Even if few people today find benefit there are still untold people in the future who may find this fascinating.

[wandelaar]

30 minutes ago, Marblehead said:

Yes, I still have questions but I don't know what they are.  How's that for being insecure?

 

So yes, I can plot a point on a graph -1,3

 

And I can draw an arrow from point 0,0 to -1,3 and define it as arrow -1,3i.

 

But if you ask me what I have done all I can say is that I drew an arrow from 0,0 to -1,3 and called that arrow -1,3i.

 

One minor correction: the arrow is written as -1 + 3i , not as -1,3i . But apart from that I think you understand all there is to understand at this point. All we have done is naming arrows in a Cartesian coordinate system in a certain way.

 

The meaning of all this is that we can now consider complex expressions of the form a + bi as designating arrows in a Cartesian coordinate system. So in our approach the complex numbers are the arrows. Complex numbers considered as being arrows thereby loose their aura of mysticism. Statements about those numbers become statements of geometrical facts about arrows. So we will not need any of the usual heavy algebraic machinery to introduce the complex numbers. We only need the understanding that in our approach the complex numbers are the arrows.

Edited August 29, 2018 by wandelaar

[Marblehead]

If you think we are ready then I guess we should proceed.

 

[wandelaar]

Two further definitions:

 

The x-coordinate of the endpoint of a complex number (= arrow) z is called the real part Re(z) of the complex number z .

The y-coordinate of the endpoint of a complex number (= arrow) z is called the imaginary part Im(z) of the complex number z .

 

Below I have drawn four complex numbers (= arrows) z1, z2, z3, z4 and I like you to give me the real and imaginary parts of those complex numbers.

 

[wandelaar]

Let me do z1 as an example:

Re(z1) = 2

Im(z1) = 3

Edited August 29, 2018 by wandelaar

[Marblehead]

My first response.  Holy Shit!!!

 

Okay.  I have recovered.  Back to the graph.

 

 

[Marblehead]

Here goes:

 

Z1  ->  2(real) + 3i(imaginary)

Z2  ->  3(real) + -3i(imaginary)

Z3  ->  -1(real) + -1i(Imaginary)

Z4  ->  -3(real) + 1i(imaginary)

[wandelaar]

That's almost correct. The usual mistake for beginners (and you are also falling into it) is to give the imaginary part complete with the "i", but that's not how it is defined. Our definition is:

 

Quote

The y-coordinate of the endpoint of a complex number (= arrow) z is called the imaginary part Im(z) of the complex number z

 

And the y-coordinate of a point in our Cartesian coordinate system is always a real number. That's why in the example of my previous post (regarding z1) I left out the "i" and wrote: Im(z1) = 3 .

[wandelaar]

Another test:

 

 

Can you now give the following values?

 

Re(z1) =

Im(z1) =

 

Re(z2) =

Im(z2) =

 

Re(z3) =

Im(z3) =

 

 

[Marblehead]

10 hours ago, wandelaar said:

That's almost correct. The usual mistake for beginners (and you are also falling into it) is to give the imaginary part complete with the "i", but that's not how it is defined. Our definition is:

 

 

And the y-coordinate of a point in our Cartesian coordinate system is always a real number. That's why in the example of my previous post (regarding z1) I left out the "i" and wrote: Im(z1) = 3 .

So are you saying that Z1 should have been:

 

Z1  ->  2(real) + 3(real) i(imaginary)

 

 

[wandelaar]

5 minutes ago, Marblehead said:

So are you saying that Z1 should have been:

 

Z1  ->  2(real) + 3(real) i(imaginary)

 

No - the idea is good, but your notation is weird. When a mathematician wants to say that the real part of z1 is 2 they will use the concise notation:  Re(z1) = 2 . And when a mathematician wants to say that the imaginary part of z1 is 3 they will use the concise notation:  Im(z1) = 3 . We better get this right. Can you now give the answers to my previous post using the concise notation with the symbols Re(  ) and Im(  )?

[Marblehead]

7 minutes ago, wandelaar said:

 

No - the idea is good, but your notation is weird. When a mathematician wants to say that the real part of z1 is 2 they will use the concise notation:  Re(z1) = 2 . And when a mathematician wants to say that the imaginary part of z1 is 3 they will use the concise notation:  Im(z1) = 3 . We better get this right. Can you now give the answers to my previous post using the concise notation with the symbols Re(  ) and Im(  )?

Well, I have never claimed that I am not weird.

 

Let's push the pause button and stay with "Z1" for the moment.  I feel I need to get this planted in my brain else I will not be able to go forward.

 

Please present to me how you would expect "Z1" to be displayed.  Then I will be able to ask questions if it doesn't fit into my brain.

 

 

[wandelaar]

It is not so much that I expect z1 to be displayed, but that I expect the real part Re(z1) and the imaginary part Im(z1) of z1 to be displayed. And that would look like this:

 

Re(z1) = 2

Im(z1) = 3

 

[Marblehead]

Am I hearing you correctly by thinking that you are suggesting that "3", even though it is a real number, is actually an imaginary number because or the "i"?

 

If so then:

 

X (2) (a) is a real number

 

but

 

Y (3i) (bi) is imaginary because (3 X i) is undefined. 

 

We have an X coordinate but Y could be who knows where at this point in my understanding.

 

 

 

 

[sagebrush]

math is my nemesis

although I probably could be able to figure it out better had I not experienced struggle from fourth grade on---

although I was happy to discover I enjoyed the foil method of algebra.

 

I am questioning the method or reason for your color choices on the graphing

 

why not other colors? tertiary or brown gray black silver

 

so it seems like a trap-box

 

I have to understand your formula and I do not want to--

 

 

[wandelaar]

@ Marblehead

 

No no - we are still working in the  real Cartesian coordinate system. Both the x-axis and the y-axis contain real numbers. You simply cannot construct a complex plane before you have introduced the complex numbers. That would be meaningless, because then you wouldn't have any imaginary numbers (they are not introduced yet!) to put along the y-axis. Nevertheless in technical education that's the usual way to proceed. And no one will ever notice, because as technicians they are only interested in applications. When you give a technician some new mathematical tools and tricks and show them how to work with it, they will be perfectly happy to start calculating with their new toys. And no technician will ever care about what it all means or why it works. From their practical technical viewpoint it simply doesn't matter what it means or why it works, as long as it does. And so we had all of those Bums jumping in at the start to lance you right away into the complex plane, and present you with Euler's formula before you even knew what was happening.

 

The complex plane where the y-axis contains imaginary numbers is in fact a mathematically useful device, but only after we have put the complex numbers on a solid foundation. And that foundation must naturally not contain the complex or imaginary numbers themselves. Please forget the complex plane for now, because it will only hinder the geometrical approach we are following.

[Marblehead]

Oh, My.  I got another "No,No."  Hehehe.  I'm really trying though although it may not appear so.

 

So let me see if we can define where I am in my understanding.

 

We have a graph  We have a real number X value and we have a real number Y value.  We can plot X,Y on the graph.

 

Then, just to scatter the marbles we add a "i" to the Y value so that we can call the plot on the graph a complex number and we draw an arrow from the point 0,0 to the plotted X,Y.

 

However, we did a naughty by attaching the i to the Y.  The marbles got scattered.

 

[wandelaar]

2 minutes ago, Marblehead said:

Oh, My.  I got another "No,No."  Hehehe.  I'm really trying though although it may not appear so.

 

Most people would't even have started or have given up long ago. Thus you definitely are motivated. That's: Yes yes.

 

2 minutes ago, Marblehead said:

So let me see if we can define where I am in my understanding.

 

Yes yes.

 

2 minutes ago, Marblehead said:

We have a graph  We have a real number X value and we have a real number Y value.  We can plot X,Y on the graph.

 

Yes yes.

 

2 minutes ago, Marblehead said:

Then, just to scatter the marbles we add a "i" to the Y value so that we can call the plot on the graph a complex number and we draw an arrow from the point 0,0 to the plotted X,Y.

 

Yes yes.

 

2 minutes ago, Marblehead said:

However, we did a naughty by attaching the i to the Y.  The marbles got scattered.

 

The real part and the imaginary part are defined as the coordinates of the endpoint, and those are real numbers. So now please try again to do this test:

 

3 hours ago, wandelaar said:

Another test:

 

 

Can you now give the following values?

 

Re(z1) =

Im(z1) =

 

Re(z2) =

Im(z2) =

 

Re(z3) =

Im(z3) =

 

 

 

[Marblehead]

Re(z1) = 2

Im(z1) = 0

 

Re(z2) = -1

Im(z2) = 0

 

Re(z3) = 0

Im(z3) = -2