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  1. #1 complex number division 
    Forum Masters Degree bit4bit's Avatar
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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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  3. #2 Re: complex number division 
    Forum Junior DivideByZero's Avatar
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    Quote Originally Posted by bit4bit
    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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  4. #3  
    Forum Masters Degree bit4bit's Avatar
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    Thanks, though I'm not sure what multiplying by (c<sup>2</sup>-d<sup>2</sup>)/(c<sup>2</sup>-d<sup>2</sup>) 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 i<sup>2</sup> 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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  5. #4  
    Forum Professor serpicojr's Avatar
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    He meant you're multiplying top and bottom by c-di, the conjugate of c+di. This makes the denominator c<sup>2</sup>+d<sup>2</sup>, 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)/(a<sup>2</sup>+b<sup>2</sup>)
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  6. #5  
    Forum Masters Degree bit4bit's Avatar
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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-bdi<sup>2</sup>] / [(c<sup>2</sup>-cdi+cdi-d<sup>2</sup>i<sup>2</sup>]
    = [ac+bd-adi+bci] / [c<sup>2</sup>+d<sup>2</sup>]

    = [(ac+bd) / c<sup>2</sup>+d<sup>2</sup>] + [(bci-adi) / c<sup>2</sup>+d<sup>2</sup>]

    = [(ac+bd) / c<sup>2</sup>+d<sup>2</sup>] + [(bc-ad) / c<sup>2</sup>+d<sup>2</sup>]*i
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  7. #6  
    Forum Masters Degree bit4bit's Avatar
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    I'm having some troubles now with understanding the exponential form of a complex number: z = re<sup>iφ</sup> 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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  8. #7  
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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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  9. #8  
    Forum Professor serpicojr's Avatar
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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) = <sub>n=0</sub>∑<sup>∞</sup> z<sup>n</sup>/n!

    We also know that:

    sin(z) = <sub>n=0</sub>∑<sup>∞</sup> (-1)<sup>n</sup>z<sup>2n+1</sup>/(2n+1)!

    cos(z) = <sub>n=0</sub>∑<sup>∞</sup> (-1)<sup>n</sup>z<sup>2n</sup>/(2n)!

    So letting x be a real number (although I guess this doesn't really matter just yet), we have, using -1 = i<sup>2</sup>:

    cos(x)+isin(x) =
    <sub>n=0</sub>∑<sup>∞</sup> (-1)<sup>n</sup>z<sup>2n</sup>/(2n)! + i<sub>n=0</sub>∑<sup>∞</sup> (-1)<sup>n</sup>z<sup>2n+1</sup>/(2n+1)! =
    <sub>n=0</sub>∑<sup>∞</sup> i<sup>2n</sup>z<sup>2n</sup>/(2n)! + i<sub>n=0</sub>∑<sup>∞</sup> i<sup>2n</sup>z<sup>2n+1</sup>/(2n+1)! =
    <sub>n=0</sub>∑<sup>∞</sup> i<sup>2n</sup>z<sup>2n</sup>/(2n)! + <sub>n=0</sub>∑<sup>∞</sup> i<sup>2n+1</sup>z<sup>2n+1</sup>/(2n+1)! =
    <sub>n=0</sub>∑<sup>∞</sup> i<sup>n</sup>z<sup>n</sup>/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:

    e<sup>iπ</sup> + 1 = 0

    Also note that since cos<sup>2</sup>(x)+sin<sup>2</sup>(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 R<sup>2</sup>--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(x<sup>2</sup>+y<sup>2</sup>), 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|e<sup>iϑ</sup>

    Note that we could then write this as:

    z = |Re(z),Im(z)|e<sup>i arg(z)</sup>
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  10. #9  
    Forum Masters Degree bit4bit's Avatar
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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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  11. #10  
    Forum Professor serpicojr's Avatar
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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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  12. #11  
    Forum Masters Degree bit4bit's Avatar
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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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  13. #12  
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    x' is just another real number.
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  14. #13  
    Forum Masters Degree bit4bit's Avatar
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    Quote Originally Posted by serpicojr
    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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  15. #14  
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    Quote Originally Posted by serpicojr

    2. Argument of a Complex Number

    Now we can think of the complex numbers as being the plane R<sup>2</sup>--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(x<sup>2</sup>+y<sup>2</sup>), 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|e<sup>iϑ</sup>

    Note that we could then write this as:

    z = |Re(z),Im(z)|e<sup>i arg(z)</sup>
    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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  16. #15  
    Forum Professor serpicojr's Avatar
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    Quote Originally Posted by bit4bit
    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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  17. #16  
    Forum Masters Degree bit4bit's Avatar
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    I'm trying to do it using the double angle formulae for sin and cos:

    sin(x+x') = sin(x)cos(x')+cos(x)sin(x')
    cos(x+x') = cos(x)cos(x')+sin(x)sin(x')

    ... then substituting each cos(x) with e(ix) - i*sin(x), and subtituting each sin(x) with [e(ix) - cos(x)] / i. I'm getting some really long expressions, and I'm trying to simplify them now. Am I on the right track?
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  18. #17  
    Forum Professor serpicojr's Avatar
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    You've got the right idea in using the addition formulas, but it'll help if you use the correct version for cosine:

    cos(x+x') = cos(x)cos(x')-sin(x)sin(x')

    Why not try starting with the expression e(ix)e(ix') and working back?
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  19. #18  
    Forum Masters Degree bit4bit's Avatar
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    Thanks, I don't know why I didn't try that way first, it's so much easier. Here's my working:

    e(ix)e(ix') = (cos(x)+i*sin(x))(cos(x')+i*sin(x'))

    = cos(x)cos(x')+i*cos(x)sin(x') + i*cos(x')sin(x)-sin(x)sin(x')
    = i*(cos(x)sin(x')+cos(x')sin(x)) + cos(x)cos(x')-sin(x)sin(x')
    = i*sin(x+x') + cos(x+x')
    = e(i(x+x'))

    So basically since ix and ix' are complex numbers (with real part zero) is this just showing that addition of indicies holds for complex numbers too?...i.e that a<sup>z</sup>a<sup>z'</sup>=a<sup>z+z'</sup>.
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  20. #19  
    Forum Masters Degree bit4bit's Avatar
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    Quote Originally Posted by serpicojr
    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
    Oh yeh, so e(x+iy)=e(x)e(iy) = e(x)(cos(y)+i*sin((y)), so thats what you get for a complex number with real and imaginary part.

    It satisfies the Cauchy-Riemann equations because

    f(z) = e(z) = e(x)(cos(y)+i*sin(y)) = e(x)cos(y) + ie(x)sin(y)

    So if f(x+iy) = u + iv, then u=e(x)cos(y), and v=e(x)sin(y), so:

    ∂u/∂x=e(x)cos(y) and, ∂v/∂y=e(x)cos(y), therefore, ∂u/∂x=∂v/∂y

    And:

    ∂u/∂y= -e(x)sin(y) and ∂v/∂x= e(x)sin(y), therefore, ∂u/∂y= -∂v/∂x
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  21. #20  
    Forum Professor serpicojr's Avatar
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    Booya! So now we know that:

    (d/dz)(exp(z)) = (∂u/∂x)+i(∂v/∂x)

    And noting that ∂u/∂x = u and ∂v∂x = v, we have:

    = u+iv = exp(z)

    So the complex exponential function, as we have defined it, satisfies the exponent addition formula, is holomorphic, and is its own derivative. This is what we expect, and it's really what we need! And now you can use these properties to your heart's content.
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  22. #21  
    Forum Masters Degree bit4bit's Avatar
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    Cool!

    By holomorphic, do you mean that it's continuous and differentiable?

    I've seen all sorts of terms like holomorphic, isomorphic, homeomorphic, and meromorphic functions, and never really known what they mean.

    Doesn't it have something to do with describing functions as maps, like f:R<sup>2</sup>-->C (or any functions from one set to another), and whether they are injective/surjective/bijective?
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  23. #22  
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    Holomorphic = complex differentiable = satisfies the Cauchy-Riemann equations. Differentiable implies continuous. Meromorphic is like holomorphic except it allows for "poles", places where the function is not continuous but rather goes to infinity in a controlled way. You can think of meromorphic functions as being the ratio of holomorphic functions, and dividing by 0 results in a pole.

    Isomorphisms and homeomorphisms are terms you'll first encounter in algebra and topology, respectively. They both describe bijective functions between two mathematical structures which "preserve the structure" in both directions. More generally, isomorphism is used to mean "equivalence" of mathematical structures.
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  24. #23  
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    Ok thanks.
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