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anonymous

  • 5 years ago

Ok, so now, z^5=-i How do i find the roots of this?

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  1. anonymous
    • 5 years ago
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    \[-i = e ^{-i \frac{\pi}{2}}\]

  2. anonymous
    • 5 years ago
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    \[ z^5 = e ^{-i \frac{\pi}{2} +2\pi ki}\]

  3. anonymous
    • 5 years ago
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    then factor out the i\[z^5 = e ^{i \pi [ 2k-\frac{1}{2} ] } \]

  4. anonymous
    • 5 years ago
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    take 5 root both sides\[ z= e ^{\frac{i \pi}{5} [2k -\frac{1}{2} ]}\]

  5. anonymous
    • 5 years ago
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    then sub in k=-2,-1,0,1,2 to get the five solutions

  6. anonymous
    • 5 years ago
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    \[e ^{\frac{-i \pi}{2} }\]

  7. anonymous
    • 5 years ago
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    prob cant even even see that , ahh you can do the rest

  8. anonymous
    • 5 years ago
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    just subing in numbers

  9. anonymous
    • 5 years ago
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    thanks man, EE for life

  10. anonymous
    • 5 years ago
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    how did you know to use k=-2,-1,0,1,2?

  11. anonymous
    • 5 years ago
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    kinda guessing ( but not really ) you spead the values of k as even as possible between positive and negative numbers

  12. anonymous
    • 5 years ago
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    so if you had a power of 5 you would do as I did

  13. anonymous
    • 5 years ago
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    if you had power of 3 you use k= -1,0,1

  14. anonymous
    • 5 years ago
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    if you had an even power , say 6 , then you would give an extra k to the positives , so to speak. so k=-2,-1,0,1,2,3

  15. anonymous
    • 5 years ago
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    oh ok, thanks :D

  16. anonymous
    • 5 years ago
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    nevertheless , doesnt matter what values you use , you could use any values of k you like , just that if you had say a power of 5 ( ie z^5 ) , and you chose k=5,10,15,20 ( ie multiples of the power ) , then you would find out that all the 5solutions you got would all be the same complex number , just that it has been rotated through several revolutions about the origin

  17. anonymous
    • 5 years ago
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    you have to keep going until you find n different solutions ( for the general eqn z^n = x+iy )

  18. anonymous
    • 5 years ago
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    thats why its easiet to have the values of k all following one another , and split over positive and negatives

  19. anonymous
    • 5 years ago
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    also, you should go through and make sure that the arguments of the solutions are in the range -pi<theta<= pi

  20. anonymous
    • 5 years ago
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    is \[-i=e^\pi/2\] an identity or something?

  21. anonymous
    • 5 years ago
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    -i=e^(pi/2)

  22. anonymous
    • 5 years ago
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    -i = e^(-pi/2) , probably cant see it well above.

  23. anonymous
    • 5 years ago
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    lol is it an idenitity?:P yes , its the polar form of -i -i has a distance from origin of 1 , and angle with positive real axis is -pi/2

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