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Math Help - Limit of complex function

  1. #1
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    Limit of complex function

    Hi,

    Can someone explain how to evaluate:

    \lim_{\Re (z)\to-\infty} \left|\frac{1+\frac{1}{2}z}{1-\frac{1}{2}z}\right|

    Would I need L'Hopital's rule?
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    Hi,

    Can someone explain how to evaluate:

    \lim_{\Re (z)\to-\infty} \left|\frac{1+\frac{1}{2}z}{1-\frac{1}{2}z}\right|

    Would I need L'Hopital's rule?
    I assume the imaginary part of z is remaining finite .....? \lim_{x \rightarrow - \infty} \left| \frac{1 + \frac{1}{2} x + i \frac{1}{2} y}{1 - \frac{1}{2} x - i \frac{1}{2} y} \right| = \lim_{x \rightarrow - \infty} \left| \frac{\frac{1}{x} + \frac{1}{2} + i \frac{y}{2x}}{\frac{1}{x} - \frac{1}{2} - i \frac{y}{2x}}\right|
    Last edited by mr fantastic; October 7th 2009 at 01:05 AM. Reason: Corrected a typo
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  3. #3
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    Yes, I believe the Im(z) remains finite. I'm basically trying to understand how Wikipedia found the limit on this page: Stiff equation - Wikipedia, the free encyclopedia , section "Example: The Euler and trapezoidal methods".

    So, about what you did: I assume y is the imaginary part of z? Where did the i vanish to?

    But from what you did, it looks like the limit becomes: (since the other terms vanish).

    \lim_{x\to -\infty} \left|\frac{ +\frac{1}{2} }{- \frac{1}{2} }\right| = |-1| = 1.

    Which is what wikipedia got, but is my working right?
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    Quote Originally Posted by scorpion007 View Post
    Yes, I believe the Im(z) remains finite. I'm basically trying to understand how Wikipedia found the limit on this page: Stiff equation - Wikipedia, the free encyclopedia , section "Example: The Euler and trapezoidal methods".

    So, about what you did: I assume y is the imaginary part of z? Where did the i vanish to?

    But from what you did, it looks like the limit becomes: (since the other terms vanish).

    \lim_{x\to -\infty} \left|\frac{ +\frac{1}{2} }{- \frac{1}{2} }\right| = |-1| = 1.

    Which is what wikipedia got, but is my working right?
    Well you know what they say, it's all fun amd laughter until someone loses an i ..... I made a typo which I've since fxed. What you have looks fne.
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    Quote Originally Posted by scorpion007 View Post
    Hi,

    Can someone explain how to evaluate:

    \lim_{\Re (z)\to-\infty} \left|\frac{1+\frac{1}{2}z}{1-\frac{1}{2}z}\right|

    Would I need L'Hopital's rule?
    I think it's much easier: put z = x + iy, so [1 +(1/2)z]/[1 - (1/2)z] = [2 + x + iy]/[2 - x - iy] = 1 + 2[x +iy]/[2 - x - iy] --> 1 + 2*(-1) = -1 when x = Re(z) --> -oo

    By the way, the same result we get when Re(z) --> oo

    Tonio
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  6. #6
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    tonio, thanks for the comment, but I don't quite understand how you got from:

    \frac{2 + x + iy}{2 - x - iy} = 1 + \frac{2(x +iy)}{2 - x - iy}

    (even though they are indeed equal)

    and then got 1 + 2*(-1) = -1 when x = Re(z) --> -oo
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  7. #7
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    Quote Originally Posted by scorpion007 View Post
    tonio, thanks for the comment, but I don't quite understand how you got from:

    \frac{2 + x + iy}{2 - x - iy} = 1 + \frac{2(x +iy)}{2 - x - iy}

    (even though they are indeed equal)

    and then got 1 + 2*(-1) = -1 when x = Re(z) --> -oo
    \frac{2 + x + iy}{2 - x - iy} = \frac{(2 - x - iy) + 2x + 2iy}{2 - x - iy} = \frac{(2 - x - iy)}{2 - x - iy} + \frac{2x + 2iy}{2 - x - iy}
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    Quote Originally Posted by scorpion007 View Post
    tonio, thanks for the comment, but I don't quite understand how you got from:

    \frac{2 + x + iy}{2 - x - iy} = 1 + \frac{2(x +iy)}{2 - x - iy}

    (even though they are indeed equal)

    and then got 1 + 2*(-1) = -1 when x = Re(z) --> -oo

    it's a basic algebra thing: if we have a quotient Q/P, where Q, P are polynomials on some unknow t of the same degree (on t), then we can always write Q/P = 1 - Q'/P, for some pol. in t of degree at most deg(Q)

    After that all is simple limits: as it is x --> -oo all the rest is taken as constants, and thus (x + constant)/(-x + constant) --> -1

    Tonio
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