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henpen Group Title

A complex question

  • 2 years ago
  • 2 years ago

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  1. henpen Group Title
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    \[\left| z-1 \right|+\left| z+1 \right|=4\] Which turns into the messy \[\sqrt{(a-1)^2+b^2}+\sqrt{(a+1)^2+b^2}=4\] Which I have no idea about how to solve graphically.

    • 2 years ago
  2. henpen Group Title
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    (Where z=a+ib)

    • 2 years ago
  3. henpen Group Title
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    (And by 'graphically' I mean plotting the possible values of z on the complex plane)

    • 2 years ago
  4. henpen Group Title
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    http://www.wolframalpha.com/input/?i=sqrt%28%28x%2B1%29%5E2%2By%5E2%29%2Bsqrt%28%28x-1%29%5E2%2By%5E2%29%3D4 Can anyone explain why the graph is an ellipse?

    • 2 years ago
  5. baddinlol Group Title
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    OK

    • 2 years ago
  6. baddinlol Group Title
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    | a + bi - 1| + |a + bi + 1| = 4

    • 2 years ago
  7. baddinlol Group Title
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    | (a -1) + (b)i | + | (a+1) + (b)i | = 4

    • 2 years ago
  8. baddinlol Group Title
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    Now let me find the modulus which is given by sqrt( re(z)^2 + Im(z)^2)

    • 2 years ago
  9. baddinlol Group Title
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    \[\sqrt{(a +1)^2 + b^2} + \sqrt{(a-1)^2 + b^2} = 4\]

    • 2 years ago
  10. baddinlol Group Title
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    Then you should square both sides

    • 2 years ago
  11. baddinlol Group Title
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    Actually that will be hard

    • 2 years ago
  12. baddinlol Group Title
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    So move one of the square roots on the left then square both sides

    • 2 years ago
  13. baddinlol Group Title
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    \[(\sqrt{(a+1)^2 + b^2} )^2= (4 - \sqrt{(a-1)^2 + b^2})^2\]

    • 2 years ago
  14. baddinlol Group Title
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    Which will give you this

    • 2 years ago
  15. henpen Group Title
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    There's still a horrible square root sign there, though.

    • 2 years ago
  16. baddinlol Group Title
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    \[(a+1)^2 + b^2= 16 - 8\sqrt{(a-1)^2 + b^2} + (a-1)^2 + b^2\]

    • 2 years ago
  17. baddinlol Group Title
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    I will get rid of it soon

    • 2 years ago
  18. baddinlol Group Title
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    Ok now i am going to cancel some stuff and simplify this equation

    • 2 years ago
  19. baddinlol Group Title
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    \[a^2 + 2a + 1 + b^2 = 16 - 8\sqrt{(a-1)^2 + b^2} + a^2 - 2a + 1 + b^2\]

    • 2 years ago
  20. baddinlol Group Title
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    Now i can simplify further

    • 2 years ago
  21. baddinlol Group Title
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    \[4a -16 = -8\sqrt{(a-1)^2+b^2}\]

    • 2 years ago
  22. henpen Group Title
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    4a-16=-8sqrt(a^2-2a+1+b^2)

    • 2 years ago
  23. baddinlol Group Title
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    Now I am going to square both sides again

    • 2 years ago
  24. baddinlol Group Title
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    \[(4a -16)^2 = 64((a-1)^2 + b^2)\]

    • 2 years ago
  25. baddinlol Group Title
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    Finally we have no more square roots

    • 2 years ago
  26. baddinlol Group Title
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    Now i am going to expand

    • 2 years ago
  27. baddinlol Group Title
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    \[16a^2 -128a + 256 = 64 a^2 -128a + 64 + 64b^2)\]

    • 2 years ago
  28. henpen Group Title
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    it ends up as 192=48a^2+64b^2 I think

    • 2 years ago
  29. baddinlol Group Title
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    Now simlify...

    • 2 years ago
  30. baddinlol Group Title
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    SSo you get

    • 2 years ago
  31. henpen Group Title
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    That actually agrees with the wolfram link I posted earlier. Thanks very much- I had little idea how to deal with square roots efficiently before!

    • 2 years ago
  32. baddinlol Group Title
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    Final step: complete the square on a term

    • 2 years ago
  33. baddinlol Group Title
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    oops

    • 2 years ago
  34. baddinlol Group Title
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    Sorry yeah that was the answer 192 = 48a^2 +64b^2

    • 2 years ago
  35. baddinlol Group Title
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    And thats an eclipse.

    • 2 years ago
  36. baddinlol Group Title
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    No problem. =)

    • 2 years ago
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