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Yahoo!

  • 3 years ago

\[|Z+ \frac{ 9 }{ Z }| = 6\] ,Then find the Greatest value of |Z| ?

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  1. JamesJ
    • 3 years ago
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    Easiest way to proceed is the square both sides and see what you get from there

  2. JamesJ
    • 3 years ago
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    so ... waiting for you to take a step or engage in the conversation

  3. funinabox
    • 3 years ago
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    couldnt you also just say Z + 9/Z = 6, and ignore the absolute value, because you are looking for the largest value anyways then solve for Z as: Z^2 + 9 = 6z Z^2 - 6x + 9 = 0 and go from there

  4. JamesJ
    • 3 years ago
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    that gives you some of the solutions. You also need to consider the case z + 9/z = -6

  5. Yahoo!
    • 3 years ago
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    @JamesJ that would give answer as |Z| = 3.... But the answer shuld be |Z| = 3 + sqrt18

  6. Yahoo!
    • 3 years ago
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    Note: Z is a Complex Number

  7. JamesJ
    • 3 years ago
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    Squaring both sides \[ \left| z + 9/z \right|^2 = 6^2 \] i.e., \[ (z + 9/z)^2 = 36 \] as |x|^2 = x^2 for all real numbers x. Hence \[ z^2 + 18 + 81/z^2 = 36 \] and therefore \[ z^4 - 18z^2 + 81 = 0 \] Now you have a quadratic equation in z^2 which you can solve... \[ (z^2 - 9)^2 = 0 \] \[ z^2 = 9 \] \[ |z| = 3 \] yes. The other answer you just wrote down doesn't satisfy the original equation, does it?

  8. JamesJ
    • 3 years ago
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    Or is z here a complex number?

  9. Yahoo!
    • 3 years ago
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    Z is a Complex Number

  10. JamesJ
    • 3 years ago
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    If z is complex, we're going to have use a different approach.

  11. JamesJ
    • 3 years ago
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    right. So in that case, what have you tried?

  12. Yahoo!
    • 3 years ago
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    No Idea..

  13. JamesJ
    • 3 years ago
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    Write z = x +iy as usual. Now notice the original equation is equivalent to |z^2 + 9| = 6|z| Now write out those two expressions in terms of x and y and see what you get. It will be easier to square them.

  14. JamesJ
    • 3 years ago
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    i.e., write out \[ |z^2 + 9|^2 = 36 |z|^2 \] for example, \[ |z|^2 = x^2 + y^2 \] You take the next step ....

  15. Yahoo!
    • 3 years ago
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    (x^2 -y^2 + 9)^2 + (2xy)^2 = 36(x^2 + y^2)

  16. JamesJ
    • 3 years ago
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    Yes, now see if you can reduce this somehow

  17. experimentX
    • 3 years ago
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    |dw:1360344301541:dw|

  18. JamesJ
    • 3 years ago
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    Yes, it's not clear to me that we get much from the polar approach. Where do you go from there? Have you studied Lagrange multipliers btw? This problem looks a lot like a Lagrangian problem. You have a constraint in x and y given by this equation, and you want to maximize f(x,y) = x^2 + y^2

  19. Yahoo!
    • 3 years ago
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    Nope..i havent Studied that

  20. experimentX
    • 3 years ago
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    |dw:1360344512672:dw|

  21. experimentX
    • 3 years ago
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    |dw:1360344710864:dw| damn this is getting complicated.

  22. experimentX
    • 3 years ago
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    |dw:1360344933595:dw|

  23. sirm3d
    • 3 years ago
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    \[6=\left|Z+\frac{9}{Z}\right|>\left||Z|-\left|\frac{9}{Z}\right|\right|\]put \[z=re^{i\theta}\\|z|=r,\quad \left|\frac{9}{z}\right|=\frac{1}{r}\] now solve the inequality in r

  24. experimentX
    • 3 years ago
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    since we have isolated the value r as function of theta we might maximize it using simple derivatives. but this is not really suggested.

  25. sirm3d
    • 3 years ago
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    oops, that's 9/r |r-9/r|<6 r-9/r < 6 r^2 - 9 < 6r r^2 - 6r < 9 r^2 - 6r + 9 < 18 (r-3)^2 < 18 (r-3) < sqrt(18) r< 3 + sqrt(18)

  26. JamesJ
    • 3 years ago
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    very nice, yes.

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