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greysica Group Title

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

  • 2 years ago
  • 2 years ago

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  1. muhammad9t5 Group Title
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    you can solve it by rationalizing |dw:1343807584287:dw|

    • 2 years ago
  2. greysica Group Title
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    i dont get it .

    • 2 years ago
  3. muhammad9t5 Group Title
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    now multiply dominaters and nominators

    • 2 years ago
  4. Neo92 Group Title
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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..

    • 2 years ago
  5. Neo92 Group Title
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    still didnt get it??

    • 2 years ago
  6. greysica Group Title
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    100i + 10i^2/100-i^2 ?

    • 2 years ago
  7. greysica Group Title
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    what should i do to get conjugate form ?

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

    • 2 years ago
  9. waterineyes Group Title
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    \[\large i^2 = -1\]

    • 2 years ago
  10. greysica Group Title
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    i need x+iy form.. :( confuse

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

    • 2 years ago
  12. waterineyes Group Title
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    The value of \(i^2 \) is -1..

    • 2 years ago
  13. greysica Group Title
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    i=10 ?

    • 2 years ago
  14. greysica Group Title
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    or 100+1 = 101 ?

    • 2 years ago
  15. waterineyes Group Title
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    Last one is right..

    • 2 years ago
  16. waterineyes Group Title
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    Similarly do it in the Numerator,,,

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

    • 2 years ago
  18. waterineyes Group Title
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    \[100i + 10 i^2 = ??\]

    • 2 years ago
  19. greysica Group Title
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    100(-1)+10(-1)^2 = -100+10 =-90 ? like this ?

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

    • 2 years ago
  21. waterineyes Group Title
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    You can change i^2 there and not i.. remain i as such..

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

    • 2 years ago
  23. greysica Group Title
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    ohh i see , 100i -10 ?

    • 2 years ago
  24. waterineyes Group Title
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    Yes...

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

    • 2 years ago
  26. waterineyes Group Title
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    Here I have separated both.. does it look like x + iy ??

    • 2 years ago
  27. greysica Group Title
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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

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

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

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

    • 2 years ago
  31. waterineyes Group Title
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    This will not be the answer for this.. Ask him again when you will meet him..

    • 2 years ago
  32. waterineyes Group Title
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    10 + i is the conjugate for this..

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

    • 2 years ago
  34. greysica Group Title
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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

    • 2 years ago
  35. waterineyes Group Title
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    Let me check..

    • 2 years ago
  36. waterineyes Group Title
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    2 - i is right..

    • 2 years ago
  37. greysica Group Title
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    okay i will write my calculation ..

    • 2 years ago
  38. waterineyes Group Title
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    Can you show me..??

    • 2 years ago
  39. greysica Group Title
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    |dw:1343814167411:dw|

    • 2 years ago
  40. greysica Group Title
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    |dw:1343814339078:dw|

    • 2 years ago
  41. greysica Group Title
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    50/25 - i (-25/25) 50/25 + i 2+i

    • 2 years ago
  42. waterineyes Group Title
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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..

    • 2 years ago
  43. waterineyes Group Title
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    |dw:1343814575597:dw|

    • 2 years ago
  44. waterineyes Group Title
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    |dw:1343814799716:dw|

    • 2 years ago
  45. waterineyes Group Title
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    Can you once try this ?? Do calculations according to it.. Try once..

    • 2 years ago
  46. greysica Group Title
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    ok wait

    • 2 years ago
  47. waterineyes Group Title
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    Wait..

    • 2 years ago
  48. waterineyes Group Title
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    |dw:1343815363469:dw|

    • 2 years ago
  49. waterineyes Group Title
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    |dw:1343815404009:dw|

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

    • 2 years ago
  51. greysica Group Title
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    yap , the asnwer is 2+i

    • 2 years ago
  52. waterineyes Group Title
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    2 + i ??

    • 2 years ago
  53. greysica Group Title
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    ohh the formula from my book is wrong , omg ..

    • 2 years ago
  54. greysica Group Title
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    nono i mean 2-i

    • 2 years ago
  55. waterineyes Group Title
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    Is it clear to you now ??

    • 2 years ago
  56. waterineyes Group Title
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    See i show you more properly how I do this types of problems..

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

    • 2 years ago
  58. greysica Group Title
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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

    • 2 years ago
  59. greysica Group Title
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    |dw:1343816101532:dw|

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

    • 2 years ago
  61. greysica Group Title
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    |dw:1343816319564:dw|

    • 2 years ago
  62. greysica Group Title
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    like that ?

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

    • 2 years ago
  64. greysica Group Title
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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

    • 2 years ago
  65. waterineyes Group Title
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    Take your time.. Welcome dear..

    • 2 years ago
  66. greysica Group Title
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    |dw:1343817123248:dw| there is i^2

    • 2 years ago
  67. greysica Group Title
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    |dw:1343817171131:dw| and dissapear in this .. where is i^2 ?

    • 2 years ago
  68. waterineyes Group Title
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    |dw:1343817264875:dw|

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

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

    • 2 years ago
  71. waterineyes Group Title
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    Yes..

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

    • 2 years ago
  73. greysica Group Title
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    |dw:1343817500482:dw| is that correct ?

    • 2 years ago