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| bit4bit |
 Posted: Tue Apr 15, 2008 2:59 pm Post subject: complex number division |
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Forum Bachelors Degree
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Can someone show me a derivation of this definition from wikipedia about complex number division?
I'm trying to refresh my understanding of complex numbers, and the addition/subtracting/multiplication are easy, but I can't see how to obtain this definition for division.
Thanks
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| DivideByZero |
| Posted: Tue Apr 15, 2008 3:32 pm Post subject: Re: complex number division |
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| bit4bit wrote: |
Can someone show me a derivation of this definition from wikipedia about complex number division?
I'm trying to refresh my understanding of complex numbers, and the addition/subtracting/multiplication are easy, but I can't see how to obtain this definition for division.
Thanks |
well its technically not division. You're just multiplying by (c^2 - d^2)/(c^2 - d^2) to a fraction. Multiplying 1 to anything does not change it. so its like a way around... |
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| bit4bit |
| Posted: Tue Apr 15, 2008 4:34 pm Post subject: |
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Thanks, though I'm not sure what multiplying by (c2-d2)/(c2-d2) would acheive, since I'd still have a complex number on both the top and the bottom of the fraction.
I've done a bit more reading, and it seems the way to tackle complex division is by making the denominator a real number, by multiplying the top and bottom by the same (third) complex value, where the denominator cancels out to a real number....presumably having an i2 term in it. I think this third value is called a 'conjugate' of the denominator..and is one that cancels out the denominator into a real number.
I think I can probably derive the expression from that now.
Thanks
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| serpicojr |
| Posted: Tue Apr 15, 2008 6:19 pm Post subject: |
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He meant you're multiplying top and bottom by c-di, the conjugate of c+di. This makes the denominator c2+d2, and the numerator can be found by multiplying a+bi and c-di as usual. Note, in particular, that this formula says the multiplicative inverse of a+bi is:
(a-bi)/(a2+b2) |
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| bit4bit |
| Posted: Wed Apr 16, 2008 4:54 am Post subject: |
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Forum Bachelors Degree
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Thanks I've done some more reading since, and I managed to work it out:
for (a+bi)/(c+di), conjugate of c+di = c-di so
[(a+bi)*(c-di)] / [(c+di)*(c-di)]
= [ac-adi+bci-bdi2] / [(c2-cdi+cdi-d2i2]
= [ac+bd-adi+bci] / [c2+d2]
= [(ac+bd) / c2+d2] + [(bci-adi) / c2+d2]
= [(ac+bd) / c2+d2] + [(bc-ad) / c2+d2]*i
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| bit4bit |
| Posted: Wed Apr 16, 2008 9:09 am Post subject: |
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I'm having some troubles now with understanding the exponential form of a complex number: z = reiφ I get that in the complex plane you have Re(z) along one axis, and Im(z) along another, so each complex number, z =x+yi can be represented as a vector <x,y>. but where does this exponential form come from? Thanks
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| MagiMaster |
| Posted: Wed Apr 16, 2008 9:44 am Post subject: |
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e^x = 1 + x + x^2/2 + x^3/6 + x^4/24 + x^5/120 + ...
e^(ix) = 1 + (ix) + (ix)^2/2 + (ix)^3/6 + (ix)^4/24 + (ix)^5/120 +...
e^(ix) = 1 + ix - x^2/2 - ix^3/6 + x^4/24 + ix^5/120 + ...
e^(ix) = (1 - x^2/2 + x^4/24 - ...) + i*(x - x^3/6 + x^5/120 - ...)
e^(ix) = cos(x) + i*sin(x)
This is just using the Taylor series for e^x, cos(x) and sin(x) and the properties of i.
Then the polar coordinates (radius, theta) can be transformed to normal coordinates by x=radius*cos(theta), y=radius*sin(theta). (x, y) is x+i*y on the complex plane, so (radius, theta) becomes radius*cos(theta) + i*radius*sin(theta) = radius*(cos(theta) + i*sin(theta) = radius*e^(i*theta). This can also be written as e^(ln(radius) + i*theta).
(Hopefully that's readable.  ) |
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| serpicojr |
| Posted: Wed Apr 16, 2008 9:49 am Post subject: |
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Okay, I'm going to do a self-contained thing on this:
1. Complex Exponential, Euler's Formula
So you must have learned the power series for the exponential for this to make sense. We have:
exp(z) = n=0∑∞ zn/n!
We also know that:
sin(z) = n=0∑∞ (-1)nz2n+1/(2n+1)!
cos(z) = n=0∑∞ (-1)nz2n/(2n)!
So letting x be a real number (although I guess this doesn't really matter just yet), we have, using -1 = i2:
cos(x)+isin(x) =
n=0∑∞ (-1)nz2n/(2n)! + in=0∑∞ (-1)nz2n+1/(2n+1)! =
n=0∑∞ i2nz2n/(2n)! + in=0∑∞ i2nz2n+1/(2n+1)! =
n=0∑∞ i2nz2n/(2n)! + n=0∑∞ i2n+1z2n+1/(2n+1)! =
n=0∑∞ inzn/n!
= exp(ix)
This famous statement, cos(x)+isin(x) = exp(ix), is called Euler's Formula. For x = π, we get Euler's Identity:
eiπ + 1 = 0
Also note that since cos2(x)+sin2(x) = 1, any number of the form exp(ix) for x a real number has modulus 1. So the numbers exp(ix) parameterize the unit circle in the complex plane.
2. Argument of a Complex Number
Now we can think of the complex numbers as being the plane R2--in fact, we can think of them as being equal as R-vector spaces. Any complex number is then a vector in the plane, with the number z = x+iy corresponding to (x,y). The magnitude of this vector is sqrt(x2+y2), and I'll denote this as |x,y| for simplicity. Now let 0 ≤ ϑ < 2π be the angle made by starting at the positive x-axis and going counterclockwise to the vector (x,y). We call ϑ the argument of z, and we denote this by arg(z). Then we have:
(x,y) = (|x,y|cos(ϑ),|x,y|sin(ϑ))
Thus:
x+iy = |x,y|cos(ϑ)+i|x,y|sin(ϑ) = |x,y|(cos(ϑ)+isin(ϑ)) = |x,y|eiϑ
Note that we could then write this as:
z = |Re(z),Im(z)|ei arg(z) |
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| bit4bit |
| Posted: Wed Apr 16, 2008 10:19 am Post subject: |
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Thanks for the replies, I've never learnt taylor series/power series of exp/trig functions, argument of a complex numbers, or Eulers identity, but I have heard the expressions before. I think I'm gonna have to read on them and i'll get back to you later. Thanks.
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| serpicojr |
| Posted: Wed Apr 16, 2008 4:40 pm Post subject: |
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Well, for now, you may as well take exp(ix) = cos(x)+isin(x) as a definition. See if you can show that:
exp(i(x+x')) = exp(ix)exp(ix')
given this definition. We can then define the complex eponential in general as:
exp(z) = exp(x+iy) = exp(x)(cos(y)+isin(y))
See if you can show this satisfies the Cauchy-Riemann equations, and then show that it satisfies (d/dz)exp(z) = exp(z). Then this is a pretty good argument that exp(z) should be the complex exponential:it satisfies all of the usual properties of the exponential and takes complex values.
I was under the impression you had covered calculus II material before. If you have but didn't see power series, it shouldn't be that big of a deal--all of the power series stuff you need is probably introduced in your book. But you may want to review sequences and series from a calc II perspective. |
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| bit4bit |
| Posted: Thu Apr 17, 2008 10:30 am Post subject: |
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Well, when I learned calculus it wasn't classified by Calculus I, II, and III, and I'm not sure if this is what is a standard syllabus in the States, but here, calculus courses differ somewhat in the material that is taught, and some are more rigorous than others. We didn't cover power series, taylor series, or get into too much analysis, only the basics of sequences, series, convergence and limits. As I said in another post, all these concepts are a bit rusty for me anyway, since I haven't used them once while applying calculus to any specific problems, so I'm revising it all now....I've been working through problems today, and its slowly coming back.
I'm really intersted in learning a bit more about all this though, particularly since signal theory in electronic enginnering is based on the use of complex numbers...(apparently this is because complex numbers take account of both amplitude and phase of a signal)
As for your questions, I want to give them a try, but I need some more revision first. Can you tell me what you mean by x'. If this is indicating a derivative then what derivative is it indicating? Thanks
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| serpicojr |
| Posted: Thu Apr 17, 2008 2:36 pm Post subject: |
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| x' is just another real number. |
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| bit4bit |
| Posted: Thu Apr 17, 2008 5:08 pm Post subject: |
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| serpicojr wrote: |
Well, for now, you may as well take exp(ix) = cos(x)+isin(x) as a definition. See if you can show that:
exp(i(x+x')) = exp(ix)exp(ix')
given this definition. |
Ok, can't you just show that
exp(i(x+x')) = exp(ix+ix') = exp(ix)exp(ix')
Without having to use the defintion
exp(ix) = cos(x)+isin(x)
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| bit4bit |
| Posted: Thu Apr 17, 2008 6:19 pm Post subject: |
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| serpicojr wrote: |
2. Argument of a Complex Number
Now we can think of the complex numbers as being the plane R2--in fact, we can think of them as being equal as R-vector spaces. Any complex number is then a vector in the plane, with the number z = x+iy corresponding to (x,y). The magnitude of this vector is sqrt(x2+y2), and I'll denote this as |x,y| for simplicity. Now let 0 ≤ ϑ < 2π be the angle made by starting at the positive x-axis and going counterclockwise to the vector (x,y). We call ϑ the argument of z, and we denote this by arg(z). Then we have:
(x,y) = (|x,y|cos(ϑ),|x,y|sin(ϑ))
Thus:
x+iy = |x,y|cos(ϑ)+i|x,y|sin(ϑ) = |x,y|(cos(ϑ)+isin(ϑ)) = |x,y|eiϑ
Note that we could then write this as:
z = |Re(z),Im(z)|ei arg(z) |
OK, I understand that actually, it's just converting from cartesian to polar and then using the identity e(ix)=cos(x) + isin(x), to get the exponential expression of the complex number. Thanks alot!! I'm still trying to read through the power/taylor series though, since I feel like I'm missing out.
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| serpicojr |
| Posted: Thu Apr 17, 2008 6:51 pm Post subject: |
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| bit4bit wrote: |
Ok, can't you just show that
exp(i(x+x')) = exp(ix+ix') = exp(ix)exp(ix')
Without having to use the defintion
exp(ix) = cos(x)+isin(x) |
You can't not use the definition because that's the only thing you have to work with. It looks to me like you're assuming that since exp(x+x') = exp(x)exp(x') for real values of x and x', then this holds for any complex numbers x and x'. With the tools at hand, you can't assume this. |
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