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IrishBoy123

  • one year ago

z = complex

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  1. IrishBoy123
    • one year ago
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    \(|z| = 2\) \(|z^2 - 4z - 3| \ge ????\)

  2. ganeshie8
    • one year ago
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    so \(z\) is on a circle of radius \(2\)

  3. IrishBoy123
    • one year ago
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    yes

  4. IrishBoy123
    • one year ago
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    i have used the Triangle Inequality 1/ looking for tips on execution 2/ looking for other approaches, especially drawing it i get >= 1

  5. ganeshie8
    • one year ago
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    im getting 7 as the minimum value

  6. IrishBoy123
    • one year ago
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    shall i post what i did?

  7. mathmate
    • one year ago
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    if |z|=2, then z^2=4, so z^2-4z-3=1-4z. where z is the circle of radius 2.|dw:1441022800703:dw| So I got -4+1=-3 as the minimum.

  8. ganeshie8
    • one year ago
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    \(|z_1-z_2| \ne |z_2|-|z_2|\) in general right

  9. ganeshie8
    • one year ago
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    \(|z^2-4z-3|\ne |z^2|-|4z|-|3|\)

  10. IrishBoy123
    • one year ago
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    following....

  11. mathmate
    • one year ago
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    but |z^2-4z-3|= |z^2-3-4z|, I hope I am not wrong here.

  12. ganeshie8
    • one year ago
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    wolfram gave 7 as the minimum i really don't know how to work it haha

  13. IrishBoy123
    • one year ago
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    i used this because its backwards \[|z_1 + z_2| \ge | |z_2| - |z_2| |\] then i said \(z_1 = z^2\), and \(z_2 = (-1)(4z+3) \) and then i did that again for \(z_2\) i got \( \ge 1\) or \(\ge 7\) i may just have chosen the wrong one

  14. ganeshie8
    • one year ago
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    |z^2-4z-3| >= 4 - |4z+3| ?

  15. IrishBoy123
    • one year ago
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    so for the first bit i get \(\ge |4 - |4z + 3| |\) then \(|4z + 3| \ge \ | |4z| - |3| |= 5\) \(|4z + 3| \le |4z| + |3| = 11\)

  16. IrishBoy123
    • one year ago
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    so is final answer |4-5| or |4-11|??

  17. ganeshie8
    • one year ago
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    wow! thats really very clever!

  18. IrishBoy123
    • one year ago
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    if it's clever its because @Loser66 & @freckles taught me

  19. ganeshie8
    • one year ago
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    looks you ppl have messed with triangle inequality a lot before !

  20. Loser66
    • one year ago
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    \[|z^2 |=|z^2-4z-3+4z+3|\leq |z^2 -4z-3|+|4z+3|\leq |z^2-4z-3|+|4z|+|3|\] Pick the far left and far right, we have \[|z|^2 \leq |z^2-4z-3| +|4z|+|3|\]

  21. Loser66
    • one year ago
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    I think you can handle it from here.

  22. ganeshie8
    • one year ago
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    indeed it looks pretty

  23. ganeshie8
    • one year ago
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    wait a sec, do we have problems with signs ? don't we end up with \(\color{red}{-}7 \leq |z^2-4z-3|\) ?

  24. ganeshie8
    • one year ago
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    which is useless..

  25. Empty
    • one year ago
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    Here's my way: \[z=2e^{i\theta}\] \[z^2-4z-3 = (2e^{i \theta})^2 -4 *2e^{i \theta} -3\] \[4e^{i2\theta } -8 e^{i \theta} -3\] We want to minimize \[\sqrt{(4e^{i2\theta } -8 e^{i \theta} -3)(4e^{i2\theta } -8 e^{i \theta} -3)^*}\] Which has a minimum at the same point the square does, so let's just look at the minimum of the expression and its complex conjugate: \[f(\theta) = -24 \cos (2 \theta ) -16 \cos (\theta) +89\] \[f'(\theta)= 0 = 24 \sin(2 \theta)+ 16 \sin (\theta)\] \[\sin \theta (1+3 \cos \theta) = 0\] So here we find all the local mins, 0, \(\pi\), \(\cos^{-1}(\frac{-1}{3})\), \(-\cos^{-1}(\frac{-1}{3})\). Then we can just plug them in and see which is smallest in magnitude. Not the prettiest!

  26. Empty
    • one year ago
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    I think these 4 points are simplest to see on a picture of \(4e^{i 2 \theta} -8e^{i \theta} -3\) Reading right to left, we center ourselves at the point in the xy-plane (-3,0) and then we create a circle of radius 8 here, and noting the negative sign we rewrite for clarity: \(4e^{i 2 \theta} +8e^{i (\theta+ \pi)} -3\) So now in half of rotation we extend a little "moon" orbit at radius 4 that goes twice as fast: |dw:1441024362202:dw|

  27. Empty
    • one year ago
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    What I mean is: \(4e^{i 2 \theta} -8e^{i \theta} -3\) Think of a point at -3 as the sun which has a little earth orbiting 8 away from it counter clockwise starting at the left. Then this earth has a little moon orbiting 4 away from it which starts to the right of the earth and completes a single orbit around the earth in the time it takes the earth to go halfway around the sun.

  28. Empty
    • one year ago
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    Here enjoy this is exactly what I'm doing (and if the name Fourier means anything to you, that's what this is) https://www.youtube.com/watch?v=QVuU2YCwHjw

  29. IrishBoy123
    • one year ago
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    thank you very much @Empty i am somewhat familiar with Fourier albeit v rusty but that is extremely interesting and will be read by me for sure in detail in due course looking again at the inequality \(|z_1 + z_2 | \le |z_1| + |z_2|\) so \(|z_1| \ge |z_1 + z_2 | - |z_2|\) \(z_1 = z^2 - 4z - 3\) and \(z_1 + z_2 = z^2 \) makes \(z_2 = 4z+3\) \(|z^2 - 4z - 3| \ge |z^2 | - |4z+3| \) \( ||4z| - |3|| \le |4z+3| \le |4z| + |3|\) \( 5 \le |4z+3| \le 11\) \(|z^2 - 4z - 3| \ge 4 -11 = -7 \) or \(|z^2 - 4z - 3| \ge 4 - 5 = -1\)

  30. ganeshie8
    • one year ago
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    Awesome! somehow im stuck at proving \(||z|-|w||\le |z-w|\) from loser's work

  31. Loser66
    • one year ago
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    @ganeshie8 These are my prof's lecture note for that parts

  32. Loser66
    • one year ago
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  33. Loser66
    • one year ago
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    Sorry for attach the first page twice. :)

  34. ganeshie8
    • one year ago
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    Gotcha! \[|z-w|^2\\~\\ = |z|^2-2\Re(z\overline{w})+|w|^2\\~\\\ \ge |z|^2-2|z\overline{w}|+|w|^2\\~\\\ = |z|^2-2|z||w|+|w|^2\\~\\\ =(|z|-|w|)^2 \] therefore \(|z-w|\ge ||z|-|w||\)

  35. IrishBoy123
    • one year ago
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    i've plugged into @Empty 's eqn and the max/min are 7, 9, 10.63

  36. ganeshie8
    • one year ago
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    thanks @Loser66 that looks a lot simpler than my mess..

  37. IrishBoy123
    • one year ago
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    https://www.desmos.com/calculator/gusfoxuqmh

  38. ganeshie8
    • one year ago
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    that is so cool!

  39. IrishBoy123
    • one year ago
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    i'm quite speechless actually. thank you everyone for all the help, so much to think about ... life is good!

  40. ganeshie8
    • one year ago
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    by definition, life is always good and math is the most interesting part! yah i stole the first phrase from Preetha :)

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