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paulzord

  • 3 years ago

'complex' numbers? Suppose w6 = -64i. Determine the six values of w.

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  1. paulzord
    • 3 years ago
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    hm i have no idea how to even start this...

  2. mathgirl
    • 3 years ago
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    is this supposed to be w raised to the 6th power?

  3. paulzord
    • 3 years ago
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    ah yes it is sorry

  4. satellite73
    • 3 years ago
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    is this \[w^6=-64i\]?

  5. satellite73
    • 3 years ago
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    if so this is not so bad.

  6. paulzord
    • 3 years ago
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    yes it is \[w ^{6} = -64i\]

  7. mathgirl
    • 3 years ago
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    you just take the 6th root, right sat?

  8. satellite73
    • 3 years ago
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    hello math girl. this site is very slow for me tonight

  9. mathgirl
    • 3 years ago
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    figure out what number you multiply by itself 6 times

  10. paulzord
    • 3 years ago
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    2

  11. mathgirl
    • 3 years ago
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    that gives you 64, and then stick on i on it

  12. satellite73
    • 3 years ago
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    yeah just take the 6th roots. if working in degrees write \[-64i=64(\cos(270)+i\sin(270))\] and then take the 6ht root of 64 which is 2 and divide 270 by 6 to get 45 so first answer is \[2(\cos(45)+i\sin(45))\]

  13. mathgirl
    • 3 years ago
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    where'd the trig stuff come from?

  14. satellite73
    • 3 years ago
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    degrees means it is easier for me to divide and get whole number not write fractions in latex

  15. mathgirl
    • 3 years ago
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    i must be confused about the type of problem here

  16. paulzord
    • 3 years ago
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    how do yo know its 270?

  17. satellite73
    • 3 years ago
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    think of where -64i is in the complex plane. it is straight down 64 units, so trig form is what i wrote. you can also use \[-64i=64e^{\frac{3\pi}{2}i}\] and do it that way

  18. satellite73
    • 3 years ago
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    oh how to i know it is 270?

  19. satellite73
    • 3 years ago
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    draw the complex plane and imagine where -64i is. it is straight down 64 units, think of it as \[0-64i\]

  20. mathgirl
    • 3 years ago
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    what math class is this? i never learned this stuff

  21. satellite73
    • 3 years ago
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    you can do this as soon as you learn trig

  22. satellite73
    • 3 years ago
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    write \[a+bi\] as \[r(\cos(\theta)+i\sin(\theta)\] where \[r=\sqrt{a^2+b^2}\] and \[\tan(\frac{b}{a})=\theta\]

  23. satellite73
    • 3 years ago
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    in any case when you want the sixth roots, take the sixth root of the absolute value and divide the angle by 6

  24. abtrehearn
    • 3 years ago
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    If\[w^{6} = -64 i,\]then\[|w| = \sqrt[6]{64}= 2.\]The angle \[\theta = 3 \pi/2,\]so one of the sixth roots of -64i will be a vertex of the regular hexagon lf side length 2 centered at z = 0, and making an angle \[\theta/6 = (3 \pi/2)/6 = \pi/4.\] That vertex is at\[z = 2 \cos(\pi/4) + 2i \sin(\pi/4) = \sqrt{2} + i \sqrt{2}.\]All the sixth roots are\[2 \cos(\pi/4 + k \pi/3) + 2i \sin(\pi/4 + k \pi / 3),\]where k is an integer.

  25. satellite73
    • 3 years ago
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    so first answer is \[2(\cos(45)+i\sin(45)\] \[=2(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\] \[=\sqrt{2}+\sqrt{2}i\]

  26. satellite73
    • 3 years ago
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    to get second answer just divide circle up into 6 equal parts with one part at 45 degrees. or go around the circle again and write \[-64i=64(\cos(630)+i\sin(630))\] and now divide 630 by 6 to get 105 so second answer is \[2(\cos(105)+i\sin(105))\] etc

  27. satellite73
    • 3 years ago
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    abtrehearn answer more elegant

  28. paulzord
    • 3 years ago
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    how did you get that angle 3pi/2 when it looks like im meant to go tan 0/2

  29. satellite73
    • 3 years ago
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    here is a picture of the unit circle in the complex plane. you can see that the angle associated with i is \[\frac{3\pi }{2}\]

  30. satellite73
    • 3 years ago
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    and by the way it is \[\theta =\tan(\frac{b}{a})\] which is this case is \[\tan(\frac{-64}{0})\] which is undefined

  31. abtrehearn
    • 3 years ago
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    The angle\[3 \pi/2\]comes from plotting (0, -64) on the x-y plane and measuring counterclockwise from the positive x-axis to the point. That's three right angles, or \[3 \pi/2\]radians.

  32. paulzord
    • 3 years ago
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    oh... click

  33. paulzord
    • 3 years ago
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    no wonder the 6th root of it is pi/2?

  34. satellite73
    • 3 years ago
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    if you want to do this the slow donkey way, which may not be a bad idea the first time you see it, start with 270 degrees (i know we really should work in radians but this will make my typing easier) then divide 270 by 6 for the next one add 360 to 270 to get 630, divide by 6 to get 105

  35. satellite73
    • 3 years ago
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    then for the next one add 360 again to get 990 and divide by 6 to get 165 etc

  36. satellite73
    • 3 years ago
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    around again you get 990+360=1350 divide by 6 to get 225 around again you get 1350+360=1710 divide by 6 get 285 around again you get 1710+360=2070 divide by 6 get 345 and finally around again you get 2070+360 = 2430 divide by 6 get 405 which is the same as 45 so you are back where you started

  37. satellite73
    • 3 years ago
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    this is certainly not the elegant way to do it, but it gets the point across

  38. abtrehearn
    • 3 years ago
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    45 + 60 = 105 105 + 60 = 165 165 + 60 = 225 225 + 60 = 285 285 + 60 = 345. Those are the angles for the vertices in degrees.

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