What is the purpose of imaginary numbers and how do they work?

[NanakiXIII]

Uhh thanks, I've seen that...

Someone told me this: (a,b)/(c,d) = ((ac+bd)/(c^2+d^2), (bc-ad)i/(c^2+d^2))

Any truth to it?

 

[Gonzolo]

It is exactly what I just demonstrated to you! Look carefully. (you don't need the i in the parenthesis, it is understood that it is there in that type of notation)

 

[NanakiXIII]

It is exactly what I just demonstrated to you!

...Is that a yes?

 

[TenaliRaman]

...Is that a yes?

yes its a yes but without the i...

(a,b)/(c,d)
Multiply numerator and denominator with (c,-d)
[(a,b)(c,-d)]/[(c,d)(c,-d)]
= (ac+bd, -ad+bc)/(c^2+d^2,0)

The denominator is real and can be absorbed in the numerator and therefore,
(ac+bd, -ad+bc)/(c^2+d^2,0)
=([(ac+bd)/(c^2+d^2)],[(-ad+bc) /(c^2+d^2)])

-- AI

 

[Gonzolo]

...Is that a yes?

If you want to learn how complex numbers work, if you are smart enough to find this site and ask such an intelligent question, if you understand how to multiply them, and if you understand how to do to add \frac{3}{5}+\frac{2}{3}, then I believe you are smart enough to understand what I posted. Rewrite it carefully, doing the steps I skipped with a pencil and paper.

It is the next step in your understanding of complex numbers. When you know this, you'll be ready for other things.

Note that the first thing I did was to multiply by 1. Yes, \frac{c-id}{c-id} = 1, exactly as \frac{13}{13} = \frac{c}{c}=1. When you divide complex numbers, the first thing you have to do is multiply by 1. If the denominator is c + id, you must multiply it by c - id, this gets the i out of the denominator.

The rest is regular algebra, taking into account the definition of i. Now Hut! Hut! Hut!

 

[NanakiXIII]

Thanks, TenaliRaman.

Ok, I guess I'll take that as a compliment, Gonzolo? Please realize that I'm not a maths whiz, I'm not even extremely good at maths. So half of the formulas you post might look like complete jibberish to me. For example, the part between the first and second equal sign in your formula (sorry, I don't know how to recreate what you posted, in post #30), I'm not sure how to interpret that. My best guess:

((a+ib)/(c+id))((c-id)/(c-id))

 

[Gonzolo]

That is correct. You seem to understand multiplication such as (a+b)(c+d). If you can do that, then (a + ib)(c + id) is done exactly the same way :

(a + ib)(c + id) = ac + iad + ibc + iibd <br /> = ac + i(ad + bc) - bd <br /> = ac - bd + i(ad + bc)

Are you comfortable with that?

 

[NanakiXIII]

I can see how you got to the part after the first equal sign, but after the second, where did that '-' come from? And where did the ii from iibd disappear to?

 

[arildno]

Remember: ii=-1

 

[NanakiXIII]

Oops. Yeah. Okay, it makes sense.

 

[JasonRox]

The one thing I realized is that in high school you are so used to working with numbers and not letters.

I picked up a short book to learn Complex Numbers. You may have seen my earlier post, which is all messed. I would avoid it strongly.

 

[arildno]

The one thing I realized is that in high school you are so used to working with numbers and not letters.

.

For some unfathomable reason, teachers all over the world seem to think that students learn more of computing with "numbers" rather than familiarizing themselves with mathematical structure/algebra.
IMO, that is the major impediment to the dissemination of mathematical knowledge to the general public

 

[JasonRox]

For some unfathomable reason, teachers all over the world seem to think that students learn more of computing with "numbers" rather than familiarizing themselves with mathematical structure/algebra.
IMO, that is the major impediment to the dissemination of mathematical knowledge to the general public

Believe it or not, I enjoy using letters.

You never need a calculator. :)

 

[mathwonk]

"For some unfathomable reason, teachers all over the world seem to think that students learn more of computing with "numbers" rather than familiarizing themselves with mathematical structure/algebra.
IMO, that is the major impediment to the dissemination of mathematical knowledge to the general public"

Teachers in many schools are evaluated and paid according, not to how substantive and effective their courses are, but how popular they are with students. This seems to correlate well with the grades they give. As long as students and their parents are customers to be pleased, and they are pleased by high grades, I do not know how to insure quality control in education.

If something is difficult to teach, the answer in american schools is to remove it from the curriculum, like proofs in geometry, and all reasoning in general. Many of todays students in college do not even know what "QED" stands for, having never seen it.

It is not that teachers are deceived as to what they should teach, it is that they are not supported for it. The current "no child left behind" idiocy is making things infinitely worse. Several of the best students i have had in my life are unable to find teaching jobs today because the requirements for teachers are so stupid, that they are judged unfit to teach, because their training is so far above what is expected.

Usually the problem is that the required course is so mickey mouse that any good student has taken it in high school, and thus considered unqualified for not having taken it in college. One of my best students, who seeks to teach high school in Texas, was deemed unqualified to teach math for lack of a linear algebra course, although she has a masters degree in mechanical engineering. She had taken linear algebra as a sophomore but the law required her to take an upper level course. When she tried to enroll in the course at university of houston she was told she was overqualified and denied admission to the course! What a catch 22.

Another of my top students is still being denied the right to teach math in virginia for lack of trivial preparation courses which are so elementary (venn diagrams?) she took them in high school or junior high, although in college she did take advanced calculus and differential topology, won the mathematics department award as the best math major in the university, and then earned a PhD in biochemistry!

Our education planning is apparently being set by some of the visibly poorest educated people in the world, e.g. our prez.

 

[arildno]

Contrary to your beliefs, mathwonk; Norwegian schools are even WORSE!

 

[mathwonk]

well that is a surprize. I have not been to Oslo since 1970, and presumably things have chnaged all over the world then.

 

[arildno]

Just to give you one horrifying fact:
1) Maths education at elementary school is so cut down, that the average Norwegian pupil never start doing fractions before the age 13+

 

[mathwonk]

If we are to tell stories, here is one from an elite private school where i sent my child at a cost of several thousands of dollars a year (on top of my public school taxes).

Before starting 3rd grade I required him to learn his multiplications tables up to 12's. Upon his return from the first day of school I inquired if he had been sufficiently prepared. Only 8 years old, he began to laugh uncontrollably.

He finally regained his composure and reported that at this elite school they did the multiplication tables all year, at a rate of one number per month, and the first month was devoted to the zeroes times table.

 

[Gokul43201]

If we are to tell stories, here is one from an elite private school where i sent my child at a cost of several thousands of dollars a year (on top of my public school taxes).

Before starting 3rd grade I required him to learn his multiplications tables up to 12's. Upon his return from the first day of school I inquired if he had been sufficiently prepared. Only 8 years old, he began to laugh uncontrollably.

He finally regained his composure and reported that at this elite school they did the multiplication tables all year, at a rate of one number per month, and the first month was devoted to the zeroes times table.

This is hilarious ! Surely you're joking, Mr. mathwonk ?

 

[arildno]

Oh dear..(. or , )

I think that much of the troubles simply stems for a veiled contempt adults have for children (in particular, for their logical faculties).

The basic axioms of algebra can be illustrated by good, physical "examples" in such a way that 3-4 year olds should have no trouble understanding them.
But "Ooh, that's theory, that's too difficult for them.." rules the day; and maths becomes for 95% of the population utterly boring, meaningless, and repetitive, with no inner logic or life in it.

 

[Gokul43201]

In Ohio High Schools, they do not teach you what a voltage (potential difference) is during your regular physics. But they do cover Ohm's Law...and yet, for some (unfathomable to me) reason, they no not use the words "voltage" or "potential difference."

I don't get it

 

[JasonRox]

It's not that bad over here.

I pity you guys.

 

[Gonzolo]

So NanakiIII, my post 37 is just about everything there is to know about multiplicating complex numbers. If you understand that, you are ready to fully understand division, because dividing a complex number is actually multiplying it by 1, that is, a fraction with identical numerator and denominator.

You choose the fraction=1 such that its denominator is identical to the dividing denominator, except for the sign in front of the i. That is what I do in the first step below. Then, it's all multiplication exactly as in post 37.

\frac{a + ib}{c + id} = \frac{a + ib}{c + id} X 1 = \frac{a + ib}{c + id} X \frac{?}{?}

We choose c - id, because the - sign in front of its i is the opposite of the denominator's + in c + id. We call this the conjugate. (c - id is the conjugate of c + id and vice-versa).

When we choose this and multiply, the i's in the denominator disappear :

\frac{a + ib}{c + id} = \frac{a + ib}{c + id} X \frac{(c - id)}{(c - id)} = \frac{ac - iad + ibc - (i^2)bd}{c^2 - icd + icd - (i^2)d^2}

= \frac{ac - (-1)bd + i(bc - ad)}{c^2 -(-1)d^2} = \frac{ac + bd + i(bc - ad)}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} + i\frac{bc - ad}{c^2 + d^2}

...which was you mentionned you learned in couple form (x,y).

 

[NanakiXIII]

Okay, it took me a while, but I traced those formulas and see how you got there. But what I don't really see is why you use c-id. Why do you need the opposite of the denominator?

 

[TenaliRaman]

1/(sqrt(2)-1) = (sqrt(2)+1)
i multiplied numerator and denominator by sqrt(2)+1
but why did i multiplied by sqrt(2)+1??
bcos (sqrt(2)+1)(sqrt(2)-1) simplifies to an integer.
This is called rationalising the denominator.

A similar technique has been used above.
Bcos (c-id) (c+id) gives u a real number , we multiply numerator and denominator by the "conjugate" of the denominator.

-- AI

 

[NanakiXIII]

I still don't really understand. But the main thing is clearer now. Thanks.

 

[MiGUi]

The reason your calculator reports an error is the same that a student of ten years says that we can not do 5 - 7 because he doen't know anything about negative numbers.

 

[NanakiXIII]

...what? I think we were past that, but thanks anyway.

 

[Gonzolo]

Okay, it took me a while, but I traced those formulas and see how you got there. But what I don't really see is why you use c-id. Why do you need the opposite of the denominator?

The reason is that when you use it, all the i's in the denominators dissappear, so that what you get in the end is still in the form a + ib.

We always write a complex number with no i in the denominator, so as to immediately recognize the real part (a) and the imaginary part (b).

 

[NanakiXIII]

Thanks, Gonzolo.