I am new to complex numbers and trying to understand them and find real life conditions which could be interpreted by them "like positive one apple is gained apple and negative one apple is lost apple....so what is +1i apple and -1iapple" ....

I know that electromagnetism ,ac, wave analysis and quantum mechanics are explained by complex numbers but i can not hold them in my hand like apples...
So i had first thought that +1i may represent the situation when two persons A,B are making bets "gambling" over the result of match
And from point of view of person A : person A put -1i dollar with intermediate person ,person B put +1i dollar with the intermediate person .....
Here for person A, I consider the dollar he gives for the intermediary person -1i because if he loses, he transfers this -1i dollar to real -1 dollar ....
and the dollar given by B to the intermediary is +1i relative to A because if A wins he converts this +1i into real +1 dollar...
I do not know if my understanding is right but i continue to study complex numbers and i solve the following problem :
p(x)=-0.3x^2 +50 x-170 where p is the profit and x is the lamps produced per week.
"note here that i do not know the price of lamp or the cost of manufacturing and that the profit is nonlinear"
then I try to find number of lamps which make profit of 3000 dollars.
the solution is complex number with real part 83.3333 and imaginary part +-60.1849i.
Now i try to get profit for real part alone "1913.3333"and imaginary part alone"916.6667" then add them together but the result is 2830 not 3000.how this is possible????
Note that 2830 +170 =3000????!!!

During this thinking i tried to calculate the imaginary part after conversion to real 60.1849 ,then i thought of this positive imaginary part as taking full price in advance "down payment of 1752.5783" for future production "to get beyond the maximum profit per week of 1913.3333" and I thought of the negative imaginary part as stealing this 60 lamps from the deal and reselling them so we also get beyond barrier of weekly maximum profit...
but calculations for both cases were not equal to 3000 "1913.3333 + 1752.5783 = 3665.9117" does my guessings about meanings of imaginary part as price in advance or stealing true???

I know that my question may be vague ... this is because i am confused about meaning of complex numbers and i will accept all edits to make it more clear...

  • 2
    $\begingroup$ There's no direct analog of complex numbers as physical entities. You shouldn't be surprised, though. There's also no good analog of negative numbers, or of the number $10^{10^{100}}$, or of a lot of other abstract mathematical concepts. Complex numbers, nevertheless, are useful, since they add lots of structure to concepts like exponentials or polynomials, which can then be used to prove more general results about them. $\endgroup$ – URL Nov 9 '19 at 4:23
  • $\begingroup$ i see your point of view and i read about it in many sites....but there was this mathematician who described the area of land getting under water during sea tide as being negative area and thus if its shape is square then the side of this square will be square root of negative number....this is what encourages me to try to find real life examples of imaginary numbers till i gave up or find something which confirms that i really understand what complex numbers are for real $\endgroup$ – ahmed allam Nov 9 '19 at 4:30
  • $\begingroup$ I think the term "imaginary" can be misleading. I've looked around and perhaps this article can help you understand things better: betterexplained.com/articles/…. $\endgroup$ – jazhang Nov 9 '19 at 4:30
  • 3
    $\begingroup$ If you view a complex number as a positive magnitude (absolute vale) and an angular direction (a positive real number has angle of 0, and a negative real has angle of 180 and every non-zero complex number will have an angle between). Then multiplying multiples the magnitudes and adds the angle. I don't think any model will work unless a number represents a 2-d quantity with two different things being measured. And the best model will have one of the components is period and circular and the other is magnitudinal. $\endgroup$ – fleablood Nov 9 '19 at 4:56
  • 1
    $\begingroup$ @URL Neither of those examples you gave are abstract. Negative numbers represent balance. If someone has $0$ dollars and they owe someone $5$ dollars, then they have $-5$ dollars. $\left({10^{10}}\right)^{100}$ is a natural number, so what's the deal there? $\endgroup$ – Radial Arm Saw Jun 23 at 3:12

Actually no number really exists. The natural numbers are just a convention for counting things. Like, an apple is represent by 1, and see an apple together with another apple is represent by 2, and so on. But for example, you can't say what does means having -1 apple, or $\pi$ apples.

Basically math is about making some "arbitrary" definitions that makes things good. You can define whatever you want (if it is not self contradictory, like let $x=1$ and $x\neq 1)$, but you try to define things in such way that you get good results.

For example, think about analytic geometry. Its about describing plane geometric things. The usually $xy-$plane are useful for, for example, describing straight lines, because the equations are simple, like $y=2x$. But describing rotations on the $xy-$plane, usually, leads to really complicated equations, involing $\sin$ and $\cos$. How can we avoid this?

Think you want to describe plane geometric things, but you are more interested in some rotations. Instead of thinking about the $xy$ plane, why don't you introduce a "vector component" $w$ that rotate things? For example, if you have some vector $v$ and multiply this component, you would have the same vector $v$, but rotated by $90º$ ($\pi/2$ radians) counter-clockwise. Now see that if you have the number 1 (which, in the $xy-$ plane would be the vector $(1,0)$) and apply this component $w$, since multiplying things by 1 intuitively keeps it fixed, you have $1w=w.$ Multiply $w$ again, and you have $1ww=ww=w^{2}.$ But applying $w$ twice means rotating $180º$ ($\pi$ radians). The vector $(1,0)$ in the $xy-$plane rotated $180º$ is the vector $(-1,0),$ namely $-1$. So, you have $w^{2}=-1$.

So, the component $w$ satisfies $w^{2}=-1.$ Well, the mathematicians use to denote this component $w$ by $i$, so $i^{2}=-1.$

That is a way to understand the complex number, and my favorite one.

| cite | improve this answer | |
  • $\begingroup$ Your first sentence is very dubious. At best it is meaningless, or doesn't really exist, for what it's worth. $\endgroup$ – Allawonder Nov 9 '19 at 18:06
  • $\begingroup$ Which sentence? $\endgroup$ – Mateus Rocha Nov 9 '19 at 21:30
  • $\begingroup$ The $1$st: Actually no number really exists... $\endgroup$ – Allawonder Nov 10 '19 at 2:50

This is just an illustration of what Mateus Rocha has already said, but since it relates the usage of complex numbers to a real world problem, I thought it might help you in coming to grips with the use of complex numbers.

I work for an aircraft manufacturer, and we needed to figure out the location of the landing gear axles on an aircraft as it sits on the ground. For most of our aircraft models, the main landing gear use a trailing link configuration. This consists of a trunnion extending from the aircraft, with a bar (the "trailing link") connected to the trunnion by a pivot extending back to hold the wheel. A shock absorber (the "oleo") also connects to the trunnion to the trailing link. The location of the axle depends on how compressed the oleo is. Here is a simplified view of the geometry:

The location of the pivot and the anchor are known constants, as well as the distances from pivot to oleo attachment and axle. By taking a measurement of the oleo, we can know all sides of the triangle and can calculate the location of the attachment point, and therefore also the axle.

When I first did this calculation, I solved it using vectors, setting the origin at the Pivot. The Anchor was a known vector, and I rotated it down to the axle line by the amount needed to make the Oleo the right length.

The algebraic calculation was rather messy. But I reproduced the calculation in a number of variants of the tools we used, until at some point I noticed that a formula I was looking at was exactly like complex multiplication. Suddenly it occurred to me that I had been doing it the hard way.

If I view this as the complex plane, with $0$ at the pivot, and represent the Anchor, Attachment, and Axle as complex numbers, the rotation of Anchor to Attachment is just one complex multiplication, and converting the Attachment to the Axle is another (which in the actual calculation is not a real number, as the Pivot, Attachment, and Axle are not really in a straight line). The Oleo length is $|\text{Anchor} - \text{Attachment}|$, which made the calculation of their ratio more sensible as well.

The point is, any real-world calculation involving a complex number can be thought of instead as a calculation involving 2 real numbers. So it is never necessary to bring in complex numbers. But there are times when conceptually it is easier to work with complex numbers than pairs of real numbers (in the case of quantum mechanics, so much easier that it boggles the mind how hard it would be to express otherwise).

| cite | improve this answer | |
  • $\begingroup$ The point is, any real-world calculation involving a complex number can be thought of instead as a calculation involving 2 real numbers. this is very helpful....that is what i am trying to do exactly....i am trying to convert the imag part of lamps to real and calculate their profit and then add them to the profit of real part ...but yesterday i was missing the 3rd term 2ab"a^2+b^2+2ab"... now i am recalculating and if real profit +imag profit +3rd term got me 3000 then my understanding will be true and imag part is just a definition of condition of "to be +real or to be -real for +i and -i" $\endgroup$ – ahmed allam Nov 10 '19 at 7:49
  • $\begingroup$ I'm glad to help, but note that more often it is conceptually harder to do something in complex numbers than as a pair of real numbers. In my case, complex numbers were useful because they handily capture rotations and magnifications, as Mateus Rocha has said, and that is exactly what I needed. In your situation, I personally don't see where the properties of complex numbers would be helpful, so to me it seems like you are only adding complication. $\endgroup$ – Paul Sinclair Nov 10 '19 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.