Why do we use complex numbers

[serimc]

why do we need complex numbers

Hi
s=a+jb
2 dimensional expreesion.Why do we need to use this?In laplace transform or fourier ,we all transfer our mathematical expreesion to 2 dimensional domain.Right?

But why do we do this?What is the magic behind the curtains?In engineering we do everything with transforms.I do also.But still i don't know what a complex number is.
Can you explain what comlex numbers indicate please?
I hope i can expalin what i mean.
I am looking for your answers

 

[anomaly]

why use complex numbers

This may help some:

https://mathforum.org/library/drmath/view/53809.html

 

[DMan]

why do we need complex numbers?

If you apply nonperiodic signal to the electronic system your response will be nonperiodic and therefore you should observe it in time domian.

If you apply periodic signal to the system and you observe the response in time domain you will see that the response repeats it self-is periodic with the period of the input signal Therefore you are interested only in observing signal in the time in one period. the next period is a copy of past etc.

Let asume your signal is: f(t)=A*cos(w*t+fi)
A-amplitude
w-circular frequency
if you observe that in time domain you get periodic signal that goes to infinity. If you want to find a responce of system to this applied signal you should solve system of differencial equations->agony even for relative simple systems

But if you go about it in a sneaky way and go from time domain to freqency domain with the help of Euler relation witch introduces complex numbers:
e^(j*w*t)=cos(w*z)+i*sin(w*t) you find solution for the circuits response in much easier way.
your signal f(t)=A*e^(i*w*t+fi)
You should take in account that this analisys is only suitable for real, linear systems in sinusoidal steady state-transient response is over

 

[serimc]

why we use complex numbers

Thanks friends for answers.
But especially i want to understand the complex numbers role in laplace transform.
Why do we multply function with e^(-st) and then take its integral.Here s is a+jb.

When we pass a new domain ; we can use algebric operations.
But what is happening?
Do we transform our expression from 2 dimension to 3 dimension or 1dimension to 2 dimension?

(Maybe my post's title must be complex numbers or complex expression in laplace transform.)
Thanks again...

 

[DMan]

why do we need complex number

Laplace transformation is just one of many integral transformations. Most of integral transformations have only math-theoretical use. Laplace and Fourier transformation have found aplicative use and are most frequently used in physics and electronics of all the integral trasformations.

Integral transformations are only mathemathical tools we find handy. They do not carry any physical meaning. It's the same as with complex numbers. They are just mathematical tool that make our lives easier and complicated calculation easier.
Integral transformations are mostly used to solve differential equations, usual and partial. These are exactly the equations we get if try to solve the periodic systems in time domain.

Don't bother why complex part is there, the reason is mathemathical and not electrical.

 

[teteamigo]

use of complex functions apply

Complex Numbers born after fundamental algebra theorem, before this are useless.

Known F(x),

F(x)=0 what are the elements of a given https://en.wikipedia.org/wiki/Algebraic_structure (set&operations in set) that satisfies the equation.

https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

By this way, roots F(x)=0 maybe Complex.


In the other side: Laplace transform is like a projection in e^-st, in specific space of functions, and integral is the scalar product, of this space.


https://en.wikipedia.org/wiki/Laplace_transform Functional analysis


When we have a pole there is singularity/discontinuity caused by a strong component of e^-Lt with L=pole.

A zero is by weak...

The question is: Because we need complex functions to analyse real functions?
(And like many problems, the problem is the question, formulate the question)


We know that some particular set (verifies existence conditions) of real functions works well with Laplace transform.

e^-st with s=α+jw ; e^-st = e^-αt*( cos(wt)+j*sen(wt) ) is like a base

2 degrees of freedom - real and imaginary. There is 2 in 1.

For function with sinusoids (sine or cosine), and exponencials is trivial that Laplace transform analysis, is a mean to identify its presence in a function.


But there is other functions that the Laplace Transform may analyse.


This is like taylor series, all polinomial have a trivial taylor series. But taylor series exists for all differentiable functions (taylor series existence condition).



But Laplace Transform have great importance in operator calculus, Heavside beginned with substitution (empirical, at first sight works well!!!) of derivatives of an given order by a factor with same number in power, in differential equations.

 

[jhbbunch]

why do we need complex numbers

The idea behind complex numbers is just common sense and is really a matter of bookkeeping. In any physical system you choose to model - electrical, mechanical, fluid, or a compbination of the above - you have purely resistive elements that dissipate enery from a system (that is, it's gone forever from the system) and you have elements (inductive or capacitive) that store energy under certain conditions and release that energy back into the system under other certain conditions. It just so happens that this is all done with a precise phase angle of the various elements. Resistive elements (pure load) do not generate a phase change, purely inductive loads create a 90 degree phase change and purely capacitive elements create a -90 degree phase change. All of this can be beautifully and accurately described in complex math, where resistive elements go along the x-axis, inductive elements up the y-axis and capacitive elements go down the y-axis and you use basic trigonometry to find the resultants (net phase angles and magnitudes) of a system. By they way, all numbers are complex numbers, a real number is just the special instance where there are no inductive or capacitive elements present (the imaginary part equals zero).

 

[electricpete]

why we use complex functions

Complex exponentials are the eigenfunctions of LTI systems.

 

[nwo4life]

+why learn about complex numbers

Because life is unfair

 

[ptuser_eda]

physical meaning of complex numbers

It's a way that man find top solve some kind of ptoblems that he had in some time in the past We (humans) always create a new thing when we arrived to a point that aren't more reasonable explainaation

 

[pmonon]

why do we use laplace transform

Because of consistency and convenience

 

[jeremymcgowan]

why do we need complex number?

I've been told that it makes calculations easier. Complex numbers are something I think you just get you use to as a tool. It's kinda funny that such weird numbers would serve such a practical use.

 

[pmonon]

use of complex numbers

yeah..interesting...I read that it was an outcome of a whim of mathematician Cardano, who used to be a gambler and drunkard... see, what imagination and passion can give...man, dont just study all the time...go for party...girls...imagination...

 

[Jone]

why use complex number

Complex numbers aren't really "complex". Imagine that ONLY the rational numbers where invented and your are to solve the equation x^2=2. So in the rational system this equation has no solution. But if we extend our number system and mnemonically write the solution to this equation as sqrt(2), then we say that we have solved the equation. Now imagine only the real numbers existed and we are to solve x^2=-1. Again, we'll have to extend the number system to solve this equation. We mnemonically write the solution of this equation as x = +/-i, where i=sqrt(-1). When we do calculations where sqrt(2) or i are involved these numbers stands out like a sore tumb and we just calculate "around" them. My point here was to explain that the i isn't anything more "complex" than sqrt(2).

Ok, now to the transforms. Recall that the Fourierseries is a projection of a function into an orthogonal basis exp(inwt), where n=-inf to + inf. This basis is orthogonal and has certain other nice features like that the derivative of it is inw*exp(inwt). But the Fourier series accepted only peridic functions and now we also want to obtain a fourier series for a non-periodic function. But in a sense the period of non-periodic function is infinite, so lets use that and extend the fourier series. Doing so shows that Fourier series x(t) = sum(Cn*exp(inwt)) (summing from -inf to +inf, and Cn are the coefficients) will turn in to the fourier integral. You'll find this derivation in some book on this subject. So, that's where the Fourier integral comes from. But why did we choose this basis? Say, we have an LTI system with the frequency respone H(jw) and we apply the input signal x(t) = A*exp(iwt). What will the output signal look like? Because of LTI, convolution will give y(t) = H(jw)*x(t) and exp(iwt) is an eigenfunction. So the output is just the input signal times the frequency respone of the system. But the input signals doesn't usually look like A*exp(iwt)? The thing is, that if we can tranform the input signals to the form A *exp(iwt), then one can easily calculate the output signals. This is why transforms are useful in signal processing. And, the Laplace transform is similiar to the Fourier transform. Actually the Fourier transform is just a special case of the Laplace transform where the real part of s=sigma+iw is set to zero.

Jones

 

[nijirazdan]

why we use complex number

complex no is special for the reason that it has 2 components.one real n other imaginary
again confusing,whats this real n imaginary stuff???
well in physics real is the one with reference phase n the imaginary is the 90 degree phase shifted version of real part.i hav given u enough of hint,rest u find out on ur own by delving urself into the subject.try to extract5 the physical interpretation from what i hav described above,i m sure u will get it.if u do,do let me no

 

[Obi]

do we use complex numbers

the reality has the fourh dimension - time... and it is complex...
Also you may ask "why do we need negative numbers?" (in ancient times peopke didn't know about them) They make possible to substruct lager nomber from smaller one... when we have negative values this operation is legal... in such a way complex number make √-1 operation possible...

 

[PaulHolland]

importance of convolution

Complex numbers aren't really "complex". Imagine that ONLY the rational numbers where invented and your are to solve the equation x^2=2. So in the rational system this equation has no solution. But if we extend our number system and mnemonically write the solution to this equation as sqrt(2), then we say that we have solved the equation. Now imagine only the real numbers existed and we are to solve x^2=-1. Again, we'll have to extend the number system to solve this equation. We mnemonically write the solution of this equation as x = +/-i, where i=sqrt(-1). When we do calculations where sqrt(2) or i are involved these numbers stands out like a sore tumb and we just calculate "around" them. My point here was to explain that the i isn't anything more "complex" than sqrt(2).

Ok, now to the transforms. Recall that the Fourierseries is a projection of a function into an orthogonal basis exp(inwt), where n=-inf to + inf. This basis is orthogonal and has certain other nice features like that the derivative of it is inw*exp(inwt). But the Fourier series accepted only peridic functions and now we also want to obtain a fourier series for a non-periodic function. But in a sense the period of non-periodic function is infinite, so lets use that and extend the fourier series. Doing so shows that Fourier series x(t) = sum(Cn*exp(inwt)) (summing from -inf to +inf, and Cn are the coefficients) will turn in to the fourier integral. You'll find this derivation in some book on this subject. So, that's where the Fourier integral comes from. But why did we choose this basis? Say, we have an LTI system with the frequency respone H(jw) and we apply the input signal x(t) = A*exp(iwt). What will the output signal look like? Because of LTI, convolution will give y(t) = H(jw)*x(t) and exp(iwt) is an eigenfunction. So the output is just the input signal times the frequency respone of the system. But the input signals doesn't usually look like A*exp(iwt)? The thing is, that if we can tranform the input signals to the form A *exp(iwt), then one can easily calculate the output signals. This is why transforms are useful in signal processing. And, the Laplace transform is similiar to the Fourier transform. Actually the Fourier transform is just a special case of the Laplace transform where the real part of s=sigma+iw is set to zero.

Jones

This is the only person on this subject that has it correct. WE INVENTED COMPLEX NUMBERS to make life easy. Its simply a form of transformation like Laplace and Fourier etc...

Paul.

 

[roykyn]

i dont know its an interesting Q

 

[qiushidaren]

Yeah, we just invented many many things to make our life easier!

 

[savage67]

Try that:

you want to solve f(x)=x^2+4 equation.

You see that there is no actual solution for this equation.No actual solution means to that "If you draw this eq. in your coordinate system it will not cross over x axis" But to handle and make it usefull crossing the x axis is a must for the real world of an engineer. How make it usefull?

Try to rotate that graph by 180 degree around the point A(f(0),0) now it cross the x access and you can solve it.
Actually you force the equation to move a nonreal world and in this nonreal world you can solve this equation by using real world techniques
non real world equation is now
fn(x)=-x^2+4 (multiplying i rotares 90 degree , multiplying i^2 rotates 180 degree so) your non real world roots are 2 and -2
so
Real -------=----Non real
x^2 * i^2 = -x^2
so i =sqrt(-1)

So your real world roots are now 2i and -2i