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anonymous

  • 4 years ago

HELPP .. 10i/10-i solve in z=x+iy (compleks conjugate)

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  1. anonymous
    • 4 years ago
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    you can solve it by rationalizing |dw:1343807584287:dw|

  2. anonymous
    • 4 years ago
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    i dont get it .

  3. anonymous
    • 4 years ago
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    now multiply dominaters and nominators

  4. Neo92
    • 4 years ago
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    10i/10-i =(0+10i)/(10-i) =(0+10i)(10+i)/(10-i)(10+i) do the remaining few calculations and tell me the answer..

  5. Neo92
    • 4 years ago
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    still didnt get it??

  6. anonymous
    • 4 years ago
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    100i + 10i^2/100-i^2 ?

  7. anonymous
    • 4 years ago
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    what should i do to get conjugate form ?

  8. anonymous
    • 4 years ago
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    Denominator part is right but need to solve further.. \[100 - i^2 = ??\]

  9. anonymous
    • 4 years ago
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    \[\large i^2 = -1\]

  10. anonymous
    • 4 years ago
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    i need x+iy form.. :( confuse

  11. anonymous
    • 4 years ago
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    Yes you will get that.. Just have patience.. Tell me : \[100 - i^2 = ??\]

  12. anonymous
    • 4 years ago
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    The value of \(i^2 \) is -1..

  13. anonymous
    • 4 years ago
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    i=10 ?

  14. anonymous
    • 4 years ago
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    or 100+1 = 101 ?

  15. anonymous
    • 4 years ago
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    Last one is right..

  16. anonymous
    • 4 years ago
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    Similarly do it in the Numerator,,,

  17. Neo92
    • 4 years ago
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    @greysica i^2 = -1 therefore 100 - (-1) ==?

  18. anonymous
    • 4 years ago
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    \[100i + 10 i^2 = ??\]

  19. anonymous
    • 4 years ago
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    100(-1)+10(-1)^2 = -100+10 =-90 ? like this ?

  20. Neo92
    • 4 years ago
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    (100i - 10) use the fact that i^2 = -1

  21. anonymous
    • 4 years ago
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    You can change i^2 there and not i.. remain i as such..

  22. anonymous
    • 4 years ago
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    \[\large 100i + 10i^2 = 100i + 10(-1) = ??\]

  23. anonymous
    • 4 years ago
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    ohh i see , 100i -10 ?

  24. anonymous
    • 4 years ago
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    Yes...

  25. anonymous
    • 4 years ago
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    Now you have to find this in the form z = x + iy.. So: \[\frac{-10 + 100i}{101} \implies \frac{-10}{101} + i \frac{100}{101}\]

  26. anonymous
    • 4 years ago
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    Here I have separated both.. does it look like x + iy ??

  27. anonymous
    • 4 years ago
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    yes , i do understand . but my teacher already give me an answer for this , he said it should be 10+i and i must find the way to get 10 + i

  28. anonymous
    • 4 years ago
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    See your teacher said to you want to solve this by using \(10 + i\) Here we have done the same ..

  29. anonymous
    • 4 years ago
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    The conjugate of \(10 - i\) is \(10 + i\) So we have multiplied and divide it by the conjugate that is : \(10 + i\)

  30. anonymous
    • 4 years ago
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    i dont understand . we've got -10/101+i100/101 and my teacher said 10+i as the answer

  31. anonymous
    • 4 years ago
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    This will not be the answer for this.. Ask him again when you will meet him..

  32. anonymous
    • 4 years ago
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    10 + i is the conjugate for this..

  33. anonymous
    • 4 years ago
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    I think he just said to find conjugate so then the answer should be \(10 + i\)

  34. anonymous
    • 4 years ago
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    okay ,how about this.. 11+2i / 4+3i I do some calculation and get 2+i as the answer but my teacher said the asnwer is 2-i

  35. anonymous
    • 4 years ago
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    Let me check..

  36. anonymous
    • 4 years ago
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    2 - i is right..

  37. anonymous
    • 4 years ago
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    okay i will write my calculation ..

  38. anonymous
    • 4 years ago
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    Can you show me..??

  39. anonymous
    • 4 years ago
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    |dw:1343814167411:dw|

  40. anonymous
    • 4 years ago
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    |dw:1343814339078:dw|

  41. anonymous
    • 4 years ago
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    50/25 - i (-25/25) 50/25 + i 2+i

  42. anonymous
    • 4 years ago
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    You are doing mistake of minus.. Let me understand what you did and what the way of your teacher to solve this complex numbers problems..

  43. anonymous
    • 4 years ago
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    |dw:1343814575597:dw|

  44. anonymous
    • 4 years ago
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    |dw:1343814799716:dw|

  45. anonymous
    • 4 years ago
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    Can you once try this ?? Do calculations according to it.. Try once..

  46. anonymous
    • 4 years ago
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    ok wait

  47. anonymous
    • 4 years ago
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    Wait..

  48. anonymous
    • 4 years ago
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    |dw:1343815363469:dw|

  49. anonymous
    • 4 years ago
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    |dw:1343815404009:dw|

  50. anonymous
    • 4 years ago
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    Here x1 = 11, x2 = 4 y1 = 2 and y2 = 3

  51. anonymous
    • 4 years ago
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    yap , the asnwer is 2+i

  52. anonymous
    • 4 years ago
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    2 + i ??

  53. anonymous
    • 4 years ago
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    ohh the formula from my book is wrong , omg ..

  54. anonymous
    • 4 years ago
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    nono i mean 2-i

  55. anonymous
    • 4 years ago
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    Is it clear to you now ??

  56. anonymous
    • 4 years ago
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    See i show you more properly how I do this types of problems..

  57. anonymous
    • 4 years ago
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    \[\frac{11 + 2i}{4 + 3i} \times \frac{4 - 3i}{4 - 3i} \implies \frac{44 - 33i + 8i - 6i^2}{(4)^2 - (3i)^2} \implies \frac{44 - 25i -6(-1)}{16-9i^2}\] \[\frac{44 - 25i + 6}{16 - 9(-1)} \implies \frac{50 - 25i}{16 + 9} \implies \frac{50 - 25i}{25} \implies \frac{50}{25} - i \frac{25}{25} \implies 2 - i\]

  58. anonymous
    • 4 years ago
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    okay , thank you !! if u dont mind , i still have a little confuse , if z1/z2 = x1+iy1/x2+iy2 how about z2/2z1 ? is that means .. wait i draw

  59. anonymous
    • 4 years ago
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    |dw:1343816101532:dw|

  60. anonymous
    • 4 years ago
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    You have to multiply and divide by the conjugate.. Do you know how to find conjugate of a complex number ??

  61. anonymous
    • 4 years ago
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    |dw:1343816319564:dw|

  62. anonymous
    • 4 years ago
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    like that ?

  63. anonymous
    • 4 years ago
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    yes well done.. Need not to multiply 2.. You can simply do this: |dw:1343816404418:dw|

  64. anonymous
    • 4 years ago
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    hmm i see .. i really suck at this actually .. but i try to understand .. wait i try to do some calculation, thank you btw

  65. anonymous
    • 4 years ago
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    Take your time.. Welcome dear..

  66. anonymous
    • 4 years ago
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    |dw:1343817123248:dw| there is i^2

  67. anonymous
    • 4 years ago
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    |dw:1343817171131:dw| and dissapear in this .. where is i^2 ?

  68. anonymous
    • 4 years ago
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    |dw:1343817264875:dw|

  69. anonymous
    • 4 years ago
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    \[\large -y_1y_2(i^2) = -y_1y_2(-1) \implies \color{blue}{+y_1y_2}\]

  70. anonymous
    • 4 years ago
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    ohh I understand , so i^2 always meaning -1 in compleks form ?

  71. anonymous
    • 4 years ago
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    Yes..

  72. anonymous
    • 4 years ago
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    Remember: \[\large i = \sqrt{-1}\] So need to change \(i\).. \[\large i^2 = -1\] \[\large i^3 = -i\] \[\large i^4 = 1\]

  73. anonymous
    • 4 years ago
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    |dw:1343817500482:dw| is that correct ?