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anonymous

  • 5 years ago

complex numbers: find a, b such that (-bi)^2 = a^2 ?

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  1. anonymous
    • 5 years ago
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    a^2+b^2=0; a=m+ni b=p+qi (m+ni)^2+(p+qi)^2=0 m^2+p^2=n^2+q^2....(1) mn=pq;....(2) now choose m,n,p,q such that the conditions 1,2 satisfied simultaneously....

  2. anonymous
    • 5 years ago
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    are a and b complex numbers ?

  3. anonymous
    • 5 years ago
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    yes... its given..

  4. anonymous
    • 5 years ago
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    how did you get mn = pq

  5. anonymous
    • 5 years ago
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    sorry that will be mn=-pq.

  6. anonymous
    • 5 years ago
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    also you foiled wrong

  7. anonymous
    • 5 years ago
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    (m+ni)^2+(p+qi)^2 = m^2 + 2mni -n^2 + p^2 + 2pqi - q^2 = 0

  8. anonymous
    • 5 years ago
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    so???

  9. anonymous
    • 5 years ago
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    where is 2mni and 2pqi

  10. anonymous
    • 5 years ago
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    2mn=2pq => mn=pq

  11. anonymous
    • 5 years ago
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    in your algebra , i dont see it

  12. anonymous
    • 5 years ago
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    i have done it in single step.

  13. anonymous
    • 5 years ago
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    too fast for me

  14. anonymous
    • 5 years ago
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    sorry im doing another problem, helping student, one sec

  15. anonymous
    • 5 years ago
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    i dont follow your logic

  16. anonymous
    • 5 years ago
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    oh

  17. anonymous
    • 5 years ago
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    2mn i = -2pqi

  18. anonymous
    • 5 years ago
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    ok then mn = -pq, agreed

  19. anonymous
    • 5 years ago
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    something tells me that a,b are integers though originally

  20. anonymous
    • 5 years ago
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    btw: a and b are real

  21. anonymous
    • 5 years ago
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    the correct answer is: a=b?

  22. anonymous
    • 5 years ago
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    our guy here thinks that a and b are complex numbers , i thought this was simpler, lol

  23. anonymous
    • 5 years ago
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    the question is: find a and b (both real) such that (-bi)^2 = a^2

  24. anonymous
    • 5 years ago
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    if a, b are not complex then the relation (a)^2=(-bi)^2 does not hold.

  25. anonymous
    • 5 years ago
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    so a=b=0

  26. anonymous
    • 5 years ago
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    right

  27. anonymous
    • 5 years ago
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    see your question starts with "complex numbers". so i thought that

  28. anonymous
    • 5 years ago
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    sorry

  29. anonymous
    • 5 years ago
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    this equation only have a trivial solution..

  30. anonymous
    • 5 years ago
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    yeah, no fun

  31. anonymous
    • 5 years ago
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    why a = b = 0? is it because a^2 + b^2 = 0 + 0i?

  32. anonymous
    • 5 years ago
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    so the question is more interesting if you allow a and b to be complex , right?

  33. anonymous
    • 5 years ago
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    for real no. only a=b=0 holds

  34. anonymous
    • 5 years ago
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    ok

  35. anonymous
    • 5 years ago
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    a^2 = (-b*i)^2 = b^2*i^2 = b^2 *-1 ,

  36. anonymous
    • 5 years ago
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    can you find any such a,b other than 0 that holds your condition????????????????

  37. anonymous
    • 5 years ago
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    so a^2 = -b^2 , so a^2 + b^2 = 0,

  38. anonymous
    • 5 years ago
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    no because a^2>=0 , and b^2 >= 0

  39. anonymous
    • 5 years ago
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    see i think a,b must be considered as complex otherwise its very easy....

  40. anonymous
    • 5 years ago
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    just solve the damn equation, lol

  41. anonymous
    • 5 years ago
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    if you take them complex then also you will find a solution 0+0i along with some new solution...

  42. anonymous
    • 5 years ago
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    ahhh

  43. anonymous
    • 5 years ago
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    so this equation is equilvalent, find all a,b, such that a^2 + b^2 = 0, for a,b element of C

  44. anonymous
    • 5 years ago
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    damn...i have another typo: -(bi)^2 = a^2...therefore b^2 = a^2 - so the solution is a = b?

  45. anonymous
    • 5 years ago
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    right?

  46. anonymous
    • 5 years ago
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    noooooooooooooooooooooooooooooooooo

  47. anonymous
    • 5 years ago
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    (-bi)^2 does not mean -(bi)^2

  48. anonymous
    • 5 years ago
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    yes i know

  49. anonymous
    • 5 years ago
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    find a and b (both real) such that -(bi)^2 = a^2.......this is the correct problem statement.....sorry....

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