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  • one year ago

Simplify \(\sqrt{7+24i}\)

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  1. dan815
    • one year ago
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    how about just Re^itheta = 7+24i

  2. Empty
    • one year ago
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    I don't know

  3. dan815
    • one year ago
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    hm thats not nice

  4. dan815
    • one year ago
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    |dw:1434880616850:dw|

  5. dan815
    • one year ago
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    the angle seems to be exact in degrees

  6. Empty
    • one year ago
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    Ok wait I figured it out, but it doesn't seem natural: \[\sqrt{7+24i}=\sqrt{(a+bi)^2}\] \[7+24i = a^2-b^2 + 2abi\] now we split this up into two equations with two unknowns \[7=(a+b)(a-b)\]\[12=ab\] since 7 is positive and prime that means a-b=1 and a+b=7 so we have: a=4, b=3 \[\sqrt{7+24i} = 4+3i\] Although I guess it didn't really have to work out this way, they could have been rational numbers for all I know or something.

  7. dan815
    • one year ago
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    oh thats a nice trick

  8. Empty
    • one year ago
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    Next one looks like \[\sqrt{-8-6i}\] I did the same thing basically but it looks like I'm getting two answers.

  9. Empty
    • one year ago
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    I'm getting that it's equal to both \[1-3i\] and \[-1+3i\]

  10. dan815
    • one year ago
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    there could be 2 answers

  11. dan815
    • one year ago
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    going forward and backwards

  12. dan815
    • one year ago
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    |dw:1434881253505:dw|

  13. dan815
    • one year ago
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    both of them look about right

  14. dan815
    • one year ago
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    i wonder if there is another way to do this, like how about normalizing and finding the half angle first

  15. Empty
    • one year ago
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    Makes sense since one is the negative of the other, we could just factor out like this: \[\sqrt{-8-6i}=\pm i\sqrt{8+6i}\]

  16. dan815
    • one year ago
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    and then sqrting the radius

  17. dan815
    • one year ago
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    then convert that to cartesian

  18. dan815
    • one year ago
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    or would that get messy with e and pi in that mix

  19. Empty
    • one year ago
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    I think it might, unless you are able to recognize a pythagorean triple's angle which might not be immediately apparent I think. Not sure

  20. Empty
    • one year ago
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    I think this ties in nicely with what you were saying earlier, what does the square root of this complex number mean...?

  21. dan815
    • one year ago
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    for some reason i always see this anon stalker in my questions all the time http://prntscr.com/7ji57g

  22. Empty
    • one year ago
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    Is there a visual way in which we can consider the square root of -8-6i as being 1-3i or -1+3i.

  23. dan815
    • one year ago
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    well ya, like half way from origin and position x axis , and if i see the magnitude of the radius double upto the main radius

  24. Empty
    • one year ago
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    I guess that's what you did, you drew it then rotated in either direction

  25. dan815
    • one year ago
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    yeah

  26. Empty
    • one year ago
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    and then took the square root of the length

  27. dan815
    • one year ago
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    right ya squaring i mean not doubling, i confuse those 2

  28. Empty
    • one year ago
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    Yeah I gotcha, ok here's a good one looks kinda simple though http://prntscr.com/7ji5ww

  29. dan815
    • one year ago
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    we could see they are all orthogonal or

  30. dan815
    • one year ago
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    wait we can do that thing from before, normalize all and all the vectors can be written in that e^i2pik/4 form

  31. dan815
    • one year ago
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    or just take dot products and that bs

  32. Empty
    • one year ago
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    We could subtract one of the vectors from the other 3 to make that point the origin with 3 points coming off it

  33. Empty
    • one year ago
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    then we know that one of the vectors must be 90 degrees of the other, so we kinda maybe figure that out and rotate

  34. dan815
    • one year ago
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    is there a way to rewrite all these points in a rotated dimension, like a simple way

  35. dan815
    • one year ago
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    like if i wanna just rewqrite all these vectors but with the x and y axis rorated to one of the vectors

  36. Empty
    • one year ago
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    Once we subtract one from the rest, we just plug in the two furthest ones apart into matrix and take the determinant, square root that and that's the side length, multiply that by sqrt(2) and that should be the distance to the last vector right?

  37. dan815
    • one year ago
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    oh my bad i think i am confusing this question, for some reason i was thinking they are giving us vectors..

  38. dan815
    • one year ago
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    they just gave us points ah

  39. Empty
    • one year ago
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    Points, vectors, whatever

  40. dan815
    • one year ago
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    yea this still works i guess

  41. dan815
    • one year ago
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    <p1-p2 >dot <p2-p3>

  42. dan815
    • one year ago
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    that will turn them into vectors so we can do this agian

  43. dan815
    • one year ago
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    the problem before was it cud have been like a square placed weirdly on the axis, so just dotitng their points wudnt make any sense

  44. dan815
    • one year ago
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    |dw:1434882161951:dw|

  45. Empty
    • one year ago
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    Woah once you shift it by subtracting 3i from all the points you get the three vectors: 2-2i, 4, 2+2i which is |dw:1434882136744:dw|

  46. Empty
    • one year ago
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    |dw:1434882255217:dw| Since those are the midpoints on a square, this is a square.

  47. Empty
    • one year ago
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    Their solution is ugly as crap: http://www.exampleproblems.com/wiki/index.php/CV19

  48. dan815
    • one year ago
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    |dw:1434882335886:dw|

  49. Empty
    • one year ago
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    No danny don't break space and do that infinitely otherwise we'll prove that points are really squares

  50. dan815
    • one year ago
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    oh thats gross they are writing it out as lines lol

  51. dan815
    • one year ago
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    haha

  52. Empty
    • one year ago
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    Yeah they're doing it the way I would program a computer to do it maybe, but I have a brain tyvm

  53. Empty
    • one year ago
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    maybe if they had presented it as arbitrary points or something, actually maybe we should solve the arbitrary case. I think we could do it pretty easily.

  54. IrishBoy123
    • one year ago
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    why aren't you using complex exponentials first one: \(z = 25 e^{i \theta}, tan \theta = 24/7\) \(\sqrt{z} = \pm 5 e^{i \theta/2}\)

  55. dan815
    • one year ago
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    i think a faster way if u gonna program this stuff would bne if its labelled properly p1 p2 p3 p4 find the 4 mags p1-p2, p2-p3,p3-p4,p4-p1 then p1-p2 dot p2-p3, p2-p3 dot p3-p4.... if all the mages are equal and all dots are 0 then done

  56. dan815
    • one year ago
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    this way u can break out of the loop as soon as one if it fails

  57. Empty
    • one year ago
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    Here's one way to write out the corners of a square with corners at a,b,c,d: \[a=re^{i \theta}+z\]\[b=rie^{i \theta}+z\]\[c=-re^{i \theta}+z\]\[d=-rie^{i \theta}+z\] r is the distance from the center to a corner of the square. \(\theta\) is the amount of rotation we want to give it, so it really only matters to vary it from \(0 < \theta < \pi /2\) since it'll just repeat itself and z is a complex number that translates it.

  58. Empty
    • one year ago
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    So 4 equations with 4 unknowns, r, \(\theta\) are just real numbers and z has 2 numbers since it's complex. Coolio.

  59. dan815
    • one year ago
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    yeah, that wud be nice, but we wont always get lucky

  60. Empty
    • one year ago
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    What do you mean why not?

  61. dan815
    • one year ago
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    they can give us annoying points that arent centered

  62. Empty
    • one year ago
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    They don't have to be centered for this to work

  63. dan815
    • one year ago
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    find we have to take the double diagonal and reposition that intersection as the origin

  64. Empty
    • one year ago
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    Woah! I just saw something miraculous

  65. Empty
    • one year ago
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    If it's a square then we can add up all the 4 complex numbers to get the center of the square.

  66. dan815
    • one year ago
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    wait really

  67. Empty
    • one year ago
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    Then we subtract this from the 4 numbers to center it.

  68. dan815
    • one year ago
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    no matter where the squre is placed?

  69. Empty
    • one year ago
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    It's like vectors in equilibrium.

  70. Empty
    • one year ago
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    But the equilibrium is the center displaced

  71. Empty
    • one year ago
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    Let's try it on the example they have 2+i, 4+3i, 2+5i, 3i is a square. Add them all together and subract the sum from all of them.

  72. dan815
    • one year ago
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    |dw:1434883019314:dw|

  73. Empty
    • one year ago
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    Then we can literally square any of the numbers we have to get the other 3, after we normalize them.

  74. Empty
    • one year ago
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    |dw:1434883081383:dw|

  75. dan815
    • one year ago
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    yaa thats what i mean though, u wont get nice vertices though

  76. Empty
    • one year ago
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    watch, I'll do it.

  77. dan815
    • one year ago
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    ohh wait! ur saying additiong of these 4 vectors will give u center, i can see it kind of :O

  78. dan815
    • one year ago
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    woah thats cool, it makes sense it has to go toward the center

  79. Empty
    • one year ago
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    Yeah this is the best way for a computer to do it too for any polygon.

  80. dan815
    • one year ago
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    the farther away from center u are the more u are pulled to the otehr vertices

  81. Empty
    • one year ago
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    Add up all the vertices, subtract this number from all of them, normalize them, and take one arbitrary one and square it until you get all the way around the polygon. If you don't hit one then you you just pick another one cause you happened to pick a root of 1 that wasn't relatively prime or whatever.

  82. dan815
    • one year ago
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    this gives us a much simpler way of solving the square question now

  83. dan815
    • one year ago
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    this works for all regular polygons

  84. Empty
    • one year ago
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    Yeah

  85. Empty
    • one year ago
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    Because all regular n-gons are just the n-roots of 1 stretched out by some radius and then translated in the plane somewhere from this perspective essentially.

  86. Empty
    • one year ago
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    Wait this doesn't quite work hahaha

  87. Empty
    • one year ago
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    It should be the average of the points not just the sum of the points.

  88. dan815
    • one year ago
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    ya i was just looking at that lol

  89. dan815
    • one year ago
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    |dw:1434883586373:dw|

  90. Empty
    • one year ago
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    \[\frac{(2+i)+(4+3i)+(2+5i)+3i}{4} = 2+3i\] Subtract that from each of those same 4 points: -2i, 2, 2i, -2.

  91. Empty
    • one year ago
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    What could be sexier?

  92. dan815
    • one year ago
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    |dw:1434883777297:dw|

  93. dan815
    • one year ago
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    like thats an easy one to see that fails for sure

  94. Empty
    • one year ago
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    No, that doesn't fail

  95. dan815
    • one year ago
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    why not

  96. Empty
    • one year ago
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    \[\frac{0+2-2i+(2-2i)}{4}=1+i\]

  97. dan815
    • one year ago
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    oh ya i mean not this, the one wiht just adding

  98. Empty
    • one year ago
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    Oh yeah, we're over that now, the thing that makes it clear is just imagine finding the center point of each dimension independently, that's just the average of each independently so since they don't mix when they add their averages stay separate which is nice.

  99. dan815
    • one year ago
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    |dw:1434883967181:dw|

  100. dan815
    • one year ago
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    this is the same thing just said a bit differently

  101. dan815
    • one year ago
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    with the 4 vectors 2 at a time share a b c d

  102. Empty
    • one year ago
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    True but your method fails if the square looks like this: |dw:1434884123357:dw|

  103. dan815
    • one year ago
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    oh truee

  104. dan815
    • one year ago
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    well its something thoughj still as long as the point is somethere in the square

  105. Empty
    • one year ago
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    Actually specifically for the square I see something special we can do after we move a square like this to the center. We can instantly check by just comparing that the real and imaginary parts are all identical but with a difference of + or - or i.

  106. Empty
    • one year ago
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    |dw:1434884274366:dw|

  107. dan815
    • one year ago
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    lol

  108. Empty
    • one year ago
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    No offense....

  109. Empty
    • one year ago
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    lolol

  110. dan815
    • one year ago
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    we can also just shift the square by some point, so that the origin is located anywhere in the square

  111. Empty
    • one year ago
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    But you can see that there will be uniquely only two numbers for each of the four vectors after translating it to the center.

  112. Empty
    • one year ago
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    Yeah but dan do you not understand what I said before?

  113. dan815
    • one year ago
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    which one?

  114. Empty
    • one year ago
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    Like once we average the points no matter where it was, that will give us the center. Once we subtract this value out, it doesn't matter how the square is rotated.

  115. dan815
    • one year ago
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    oh ya i can see that one works no matter what

  116. dan815
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    im just seeing if there is a less computation way in some cases

  117. dan815
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    not that it matters lol

  118. Empty
    • one year ago
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    I think this is the least computational way unless you wanna do it their gross way lol

  119. dan815
    • one year ago
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    well like (-a+b, -c+d) is center is slightly faster than (a,c) + (a,d) + (b,d)+ (b,c)/ 4

  120. dan815
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    only works if origin is somewhere in square.. it doesnt make a difference, but i can see how maybe in computer programs.. it might add up eventually if someone knows all their calculations involve origin in square

  121. Empty
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    I don't understand how your calculation is realistic though

  122. dan815
    • one year ago
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    why not

  123. Empty
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    I'm not saying this calculation is what I'm doing, you have a misconception if you think that (a,c) + (a,d) + (b,d)+ (b,c)/ 4 is what my calculation does. There is no relationship like this, in general there is not a repetition that occurs like this, just look at the example we got our question from it was this which happened to only have repetition on only 2 entries and even that was because the square was rotated perfectly 45 degrees i believe.\[\frac{(2+i)+(4+3i)+(2+5i)+3i}{4} = 2+3i\]

  124. Empty
    • one year ago
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    In that example what is a,b,c, and d if there are coefficients 2,1,4,3,5,0 ?

  125. dan815
    • one year ago
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    ohh i see the problem okayy

  126. dan815
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    ah okay that makes sense

  127. dan815
    • one year ago
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    the rotations

  128. dan815
    • one year ago
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    ok ya i see, dont know why i was thinkign the complex numbers would be repeating we can have all unique points and that summing and dividing by 4 still works

  129. dan815
    • one year ago
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    they should allow rotation in the draw tool, that give so many more options

  130. Empty
    • one year ago
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    If the number of sides on the polygon is a multiple of 4 and they're regularly spaced then we can see that after we subtract this center vector we will have: \[k=2*\lceil \frac{(number \ of \ vertices)}{8} \rceil\] Where k is the number of unique vector components we'll have... See if you can figure this one out ;)

  131. dan815
    • one year ago
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    what do you mean reguarly spaced and subtract the center vector

  132. dan815
    • one year ago
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    |dw:1434885527859:dw|

  133. dan815
    • one year ago
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    that vector?