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Math Help - Complex Numbers questions regarding 1) sketching 2) caughy-riemann equation

  1. #1
    Junior Member rubix's Avatar
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    Complex Numbers questions regarding 1) sketching 2) caughy-riemann equation

    1) how to graph |z+6| >= |z|

    2) f(z) = x^2 + i y^2

    assume Cauchy-Riemann equation

    ux = vy and uy = -vx

    are satisfied for f(z)

    where z = x + iy and u(x,y) = x and v(x,y) = y

    find all values of z which satisfy C-R on f(z)

    ---

    I've can sketch |z=4| and |z| individually but how to put them together and which portion to shade?

    ---

    I started 2 with

    u(x,y) = x^2 and v(x,y) = y^2
    ux = 2x and uy = 2y
    so 2x = 2y => x = y
    also uy = - vx
    so 0 = 0

    hence, z = x+iy = x + ix = y + iy

    but when i put x = y in f(z) i get

    f(z) = x^2 + ix^2 = y^2 + iy^2

    but this f(z) does not satisfy C-R equations.
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  2. #2
    A Plied Mathematician
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    With regard to #1, I would plug in z = x + iy, take the magnitudes, and see what you get. Geometrically, you're trying to find all z such that its distance from -6 is greater than its own magnitude.

    With regard to #2, I could be wrong, but it seems to me that f(z) does satisfy the C-R equations. Remember that your x and y are no longer independent! All your computations (which look correct to me, by the way) were set up to ensure that f(z) does obey the C-R equations.
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  3. #3
    Junior Member rubix's Avatar
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    let's see, if i take -
    f(z) = x^2 + ix^2
    u = x^2 and v = x^2
    ux = vy = vx (since x = y)
    so 2x = 2x (this satisfies)
    BUT
    uy = ux = -vx = (since x = y)
    so 2x = -2x => 2 = -2 (error!!!)
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Only for the first one...

    As you know:

    z=x+iy

    and:

    |z|=\sqrt{x^2+y^2}

    In your case:

    |z+6| \geq |z|

    or:

    \sqrt{(x+6)^2+y^2}\geq \sqrt{x^2+y^2}

    Now I let you to go on with this...
    (Don't forget to separate for two cases)
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  5. #5
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    I would agree that 2=-2 is an error. However, if 2x = -2x, canceling the x's might not be the thing to do. What if you cancel the 2's? Where does that lead?
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  6. #6
    Junior Member rubix's Avatar
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    thanks but i'm aware of that method Also sprach Zarathustra. What i was looking for was more easier way...

    ex: |z-2| < 1 would be circle of radius 1 centered at 2

    likewise, |z+6| would be centered at -6 and |z| would be centered at 0. but i'm having trouble picturing both at once.
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  7. #7
    Junior Member rubix's Avatar
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    good point, x could have been 0 in that case it woudn't be appropriate to cancel them. thnx for pointing out. i'm bit slow today.
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  8. #8
    A Plied Mathematician
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    I'm slow myself. I didn't see that until I thought long and hard about your calculations. I think, in this case, you can say more than that x "could have been" 0. What do you think?
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  9. #9
    Junior Member rubix's Avatar
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    2x = -2x
    x = -x
    iff x = 0, no?

    but now that i've thought about it, x = y and x = 0 leads to trivial solution z = 0 + i 0 = 0. so there has to be another solution. i take it you have something in mind, hints please.
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  10. #10
    Junior Member slider142's Avatar
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    Quote Originally Posted by rubix View Post
    1) how to graph |z+6| >= |z|
    The first one is not as complicated as the equations may make it seem. It can be written as |z - (-6)| \geq |z - 0|. The modulus operation |...| behaves exactly like distance on the complex plane. Specifically, , letting z = x + iy = (x, y) it is the same as the set ||(x, y) - (-6, 0)|| \geq ||(x, y) - (0,0)|| on the Cartesian plane, where ||...|| is the Cartesian distance function. De-emphasizing the semantics of which plane we are on, the question is asking where on the plane the points (x, y) are closer to the origin than to the point (-6, 0). Generalizing for clarity, given two arbitrary points on a plane, A and B, where are the points that are closer to B than to A? It is easier to see if you look at the boundary of your set: where are the points that are equidistant from A and B? Note that you can then label them (-6, 0) and (0, 0), as the labels of the points did not really matter.
    While this region is not that difficult to visualize, regions that involve more than distance formulas will require you to go back to explicit calculation.
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  11. #11
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    No, I didn't have anything else in mind. I think you've shown conclusively that that definition of f(z) satisfies the C-R equations only at the origin. Keep in mind that it's entirely possible for a function to satisfy the C-R equations nowhere. I think the complex conjugate function is like that: f(x+iy)=x-iy. We would have that u_{x}=1\not=-1=v_{y}. There's nothing you can do about that one. So your function is slightly better behaved than the conjugation function, if only slightly.
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  12. #12
    Junior Member rubix's Avatar
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    thanks slider, that was insightful and something along the line of what i was looking for.

    @ Ackbeet, your argument sounds reasonable. but i'm looking at another problem now and i'm pretty much following what i did above and i'm getting even funnier result:

    given f(z) = 2x + ixy^2

    so, u(x,y) = 2x and v(x,y) = xy^2

    C-R equations:
    ux = vy
    so 2 = 2xy => 1 = xy => y = 1/x
    and uy = -vx
    so 0 = -y^2 => y = 0

    BUT, if y = 0 then how can xy yield 1??


    makes me think there probably is another approach to these problems.
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  13. #13
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    In that case, it looks like there's no region of the complex plane in which that f(z) satisfies the C-R equations.
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  14. #14
    Junior Member rubix's Avatar
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    humm i don't like the sound of that. gonna try another one, see if it yields something funny too...

    given f(z) = Im Z = y

    u = 0 v = y

    C-R equations:
    ux = vy
    so 0 = 1 ERROR!!!
    and uy = -vx
    so 0 = 0

    turns out the way i'm doing it this one does not even satisfy C-R equation. But the question is based on the assumption that it does.
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  15. #15
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    That one not satisfying the C-R equations doesn't surprise me, either. Here's one that should satisfy the C-R equations everywhere:

    f(z)=e^{z}=e^{x+iy}=e^{x}e^{iy}=e^{x}(\cos(y)+i\si  n(y)).

    If you try the C-R equations on that, I think you'll find that the entire complex plane satisfies the C-R equations for that function. Generally, because you take both the x and y partials of both u and v, most of the time, you're going to need both x and y to show up in both u and v in order to be able to satisfy the C-R equations in a sizeable portion of the complex plane.
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