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awn786

  • 3 years ago

What is the square root of minus 1. When i do it in a scientific calculator, it says 'i' (imaginary unit) and i have no idea what that means

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  1. awn786
    • 3 years ago
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    \[\sqrt{-1} = x \] what is the value of x

  2. awn786
    • 3 years ago
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    please dont put links

  3. zaynahf
    • 3 years ago
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    Imaginary numbers result when you are taking the square root of any negative number: For an example, \[\sqrt{-1}\] = i \[\sqrt{-25} = 5i\] The 'i' is kind of like the negative sign

  4. awn786
    • 3 years ago
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    so what is an imaginary unit

  5. zaynahf
    • 3 years ago
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    I know of a video you can watch if you want to

  6. awn786
    • 3 years ago
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    yes what is the vid @zaynahf

  7. hba
    • 3 years ago
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    did you see the link @awn786 It has everything.

  8. awn786
    • 3 years ago
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    yes i did but what is an imaganary unit

  9. awn786
    • 3 years ago
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    WOW OMG complicated @hba

  10. zaynahf
    • 3 years ago
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    The first link is really straight forward, try that

  11. hba
    • 3 years ago
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    @awn786 Well,I tried to simplify and put everything :)

  12. awn786
    • 3 years ago
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    thanks for trying. @hba

  13. hba
    • 3 years ago
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    You are welcome :)

  14. awn786
    • 3 years ago
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    listen and can u give a simple i word answer: \[\sqrt{-1}=x\] what is the value of x in numbers.

  15. awn786
    • 3 years ago
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    1 word answer*

  16. awn786
    • 3 years ago
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    not i word

  17. hba
    • 3 years ago
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    I can help you understand what imaginary numbers are. So,the base of imaginary numbers is actually laid upon some of the things like \[\huge \ i=\sqrt{-1}\] The other thing is \[\huge\ \omega=\frac{ -1+i \sqrt{3} }{ 2 }\]

  18. awn786
    • 3 years ago
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    whats w

  19. hba
    • 3 years ago
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    We call this omega,It's derivations is given in my link :)

  20. awn786
    • 3 years ago
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    whats w @hba

  21. hba
    • 3 years ago
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    Now as per your question. x=root{-1} x=i Here i is actually iota and it is an imaginary number.

  22. hba
    • 3 years ago
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    That's omega @awn786

  23. awn786
    • 3 years ago
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    whats omega

  24. hba
    • 3 years ago
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    @awn786 Please check the link i have provided. It has everything Please give it a read.

  25. awn786
    • 3 years ago
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    i said in numbers. your answer is basicly my question

  26. hba
    • 3 years ago
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    Well,In numbers it is sqrt{-1}

  27. awn786
    • 3 years ago
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    it should be easy shouldent it.

  28. awn786
    • 3 years ago
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    i dont understand ur link

  29. awn786
    • 3 years ago
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    what is omega

  30. hba
    • 3 years ago
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    I have already told you what omega is.

  31. awn786
    • 3 years ago
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    no u havent @hba

  32. hba
    • 3 years ago
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    I said it is -1+i sqrt{3}/2

  33. awn786
    • 3 years ago
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    ok (i dont get it but ok) i' only in year 9 doing my gcse's

  34. GCR92
    • 3 years ago
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    \[\sqrt{-1}\] doesn't exist as a the REAL number you are asking for, because there's no number that you can square (because of the SQUARE root) and gives you -1

  35. GCR92
    • 3 years ago
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    you just need to know what a square root is

  36. awn786
    • 3 years ago
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    i know what a square root is ofc

  37. GCR92
    • 3 years ago
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    that's why there's no sqrt of ANY negative numbers, because everthing squared is BIGGER than 0

  38. GCR92
    • 3 years ago
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    (possitive)

  39. awn786
    • 3 years ago
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    oh i get it so its an imaganary number because it cant exist?

  40. awn786
    • 3 years ago
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    you have the best responce

  41. GCR92
    • 3 years ago
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    EXACTLY, it CAN'T EXIST as a "normal number"

  42. awn786
    • 3 years ago
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    thank u so much.

  43. awn786
    • 3 years ago
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    ^^^^^^ @GCR92

  44. GCR92
    • 3 years ago
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    so mathematicians around XVII century had to figure out another type of numbers, called COMPLEX numbers, and the first number of complex numbers is \[\sqrt{-1}\]

  45. GCR92
    • 3 years ago
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    they INVENTED it

  46. GCR92
    • 3 years ago
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    you are welcome

  47. awn786
    • 3 years ago
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    ur a genius @GCR92

  48. GCR92
    • 3 years ago
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    not really, but thanks!

  49. GCR92
    • 3 years ago
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    ill draw you something

  50. awn786
    • 3 years ago
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    you can bring the most complex thing and make others understand. are u sure ur not related to prfsr. brian cox

  51. GCR92
    • 3 years ago
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    |dw:1358026644045:dw|

  52. GCR92
    • 3 years ago
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    hahaha you amuse me, nothing is hard when you have an amazing teacher, and I have it right now.

  53. GCR92
    • 3 years ago
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    you understand what I drew?

  54. awn786
    • 3 years ago
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    yeah u drew a number line. who is ur teacher

  55. GCR92
    • 3 years ago
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    My uncle, who's been teaching maths since he was 15 and he's now 57, anyway, let's go back to our imagination

  56. awn786
    • 3 years ago
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    |dw:1358026974868:dw|

  57. GCR92
    • 3 years ago
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    ALMOST

  58. awn786
    • 3 years ago
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    |dw:1358027044754:dw|

  59. awn786
    • 3 years ago
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    |dw:1358027175723:dw|

  60. GCR92
    • 3 years ago
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    |dw:1358027034198:dw|

  61. GCR92
    • 3 years ago
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    you put the axis pararell and it's perpendicular cutting in 0

  62. awn786
    • 3 years ago
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    |dw:1358027255837:dw|

  63. GCR92
    • 3 years ago
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    that's why i said almost, because you figured out ANOTHER line, and that would make a 300 year old mathematician proud

  64. awn786
    • 3 years ago
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    yepeee

  65. GCR92
    • 3 years ago
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    because they CAN be negative in their new axis, the imaginary axis

  66. GCR92
    • 3 years ago
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    the imaginary line

  67. GCR92
    • 3 years ago
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    it is not allowed to be a negative square root in the REAL line, but it is in the imaginary one

  68. awn786
    • 3 years ago
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    so in imaginative can they ever be 3i or always do they have to be -3i

  69. GCR92
    • 3 years ago
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    so we just call it i, like we call 1 one

  70. GCR92
    • 3 years ago
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    there are imaginary postive numbers and imaginary negative numbers, under 0

  71. GCR92
    • 3 years ago
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    real negatives are at the left of 0, so why can't there be negative imaginary ones?

  72. awn786
    • 3 years ago
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    |dw:1358027454145:dw|

  73. awn786
    • 3 years ago
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    above 0 is positive?

  74. GCR92
    • 3 years ago
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    if you delete the minus inside of the root, it would be a real number then!

  75. GCR92
    • 3 years ago
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    Yes that's what i was writting

  76. GCR92
    • 3 years ago
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    and under 0 negative, i forgot to draw it

  77. GCR92
    • 3 years ago
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    just like the real line but perpendicular

  78. GCR92
    • 3 years ago
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    above 0 goes posstive until inifinity and under 0 negative until infinity

  79. awn786
    • 3 years ago
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    and on the real line is it same?

  80. GCR92
    • 3 years ago
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    |dw:1358027632303:dw|

  81. GCR92
    • 3 years ago
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    they are two separated lines.

  82. awn786
    • 3 years ago
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    bye i need to go. if i see u tommorw here i will carry on the conversation.

  83. GCR92
    • 3 years ago
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    okay, sure! bye! you are smart btw, you figured the other line very well.

  84. GCR92
    • 3 years ago
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    figured out*

  85. awn786
    • 3 years ago
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    thanks

  86. awn786
    • 3 years ago
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    @GCR92 im back

  87. Krishnadas
    • 3 years ago
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    complex numbers

  88. AravindG
    • 3 years ago
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    i is actually an imaginary number used to give value to \[\sqrt{-1}\]

  89. awn786
    • 3 years ago
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    yes i get that.

  90. awn786
    • 3 years ago
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    Look above^

  91. ParthKohli
    • 3 years ago
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    Wait, no!

  92. awn786
    • 3 years ago
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    what?

  93. ParthKohli
    • 3 years ago
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    \(i\) is defined as the imaginary number such that \(i^2 = -1\), not \(i= \sqrt{-1}\).

  94. awn786
    • 3 years ago
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    no ur wrong

  95. awn786
    • 3 years ago
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    wait

  96. ParthKohli
    • 3 years ago
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    How so...?

  97. awn786
    • 3 years ago
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    watch

  98. awn786
    • 3 years ago
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    wait

  99. awn786
    • 3 years ago
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    http://imgur.com/kwevvqi

  100. awn786
    • 3 years ago
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    and also its just common knowlige

  101. awn786
    • 3 years ago
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    Knowledge*

  102. ParthKohli
    • 3 years ago
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    Yes, \(i\) is one of the square roots of \(-1\). But \(i \ne \sqrt{-1}\).

  103. ParthKohli
    • 3 years ago
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    I mean, \(\sqrt{-1} =i,-i\)

  104. ParthKohli
    • 3 years ago
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    So you must define \(i^2 = -1\) if you don't want things to mess up

  105. awn786
    • 3 years ago
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    wow gcr92 was WAY better at explaining things

  106. ParthKohli
    • 3 years ago
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    What would you like to learn?

  107. awn786
    • 3 years ago
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    apparantly his teacher is hgis uncle who started teaching him maths when he was 15 an now is 50

  108. awn786
    • 3 years ago
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    wait

  109. awn786
    • 3 years ago
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    carry on from:

  110. awn786
    • 3 years ago
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    |dw:1358591619083:dw|

  111. awn786
    • 3 years ago
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    look above^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

  112. ParthKohli
    • 3 years ago
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    Re and Co, yes.

  113. nikhilchabe
    • 3 years ago
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    Here you go with the link for complete solution! http://www.learnalberta.ca/content/memg/Division03/Square%20Root/index.html

  114. awn786
    • 3 years ago
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    ermmm it needs to be simple

  115. awn786
    • 3 years ago
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    nope its not

  116. awn786
    • 3 years ago
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    sorry

  117. awn786
    • 3 years ago
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    where is GCR92

  118. GCR92
    • 3 years ago
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    Here. What's up?

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