anonymous
  • anonymous
Ok, so now, z^5=-i How do i find the roots of this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[-i = e ^{-i \frac{\pi}{2}}\]
anonymous
  • anonymous
\[ z^5 = e ^{-i \frac{\pi}{2} +2\pi ki}\]
anonymous
  • anonymous
then factor out the i\[z^5 = e ^{i \pi [ 2k-\frac{1}{2} ] } \]

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anonymous
  • anonymous
take 5 root both sides\[ z= e ^{\frac{i \pi}{5} [2k -\frac{1}{2} ]}\]
anonymous
  • anonymous
then sub in k=-2,-1,0,1,2 to get the five solutions
anonymous
  • anonymous
\[e ^{\frac{-i \pi}{2} }\]
anonymous
  • anonymous
prob cant even even see that , ahh you can do the rest
anonymous
  • anonymous
just subing in numbers
anonymous
  • anonymous
thanks man, EE for life
anonymous
  • anonymous
how did you know to use k=-2,-1,0,1,2?
anonymous
  • anonymous
kinda guessing ( but not really ) you spead the values of k as even as possible between positive and negative numbers
anonymous
  • anonymous
so if you had a power of 5 you would do as I did
anonymous
  • anonymous
if you had power of 3 you use k= -1,0,1
anonymous
  • anonymous
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
anonymous
  • anonymous
oh ok, thanks :D
anonymous
  • anonymous
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
anonymous
  • anonymous
you have to keep going until you find n different solutions ( for the general eqn z^n = x+iy )
anonymous
  • anonymous
thats why its easiet to have the values of k all following one another , and split over positive and negatives
anonymous
  • anonymous
also, you should go through and make sure that the arguments of the solutions are in the range -pi
anonymous
  • anonymous
is \[-i=e^\pi/2\] an identity or something?
anonymous
  • anonymous
-i=e^(pi/2)
anonymous
  • anonymous
-i = e^(-pi/2) , probably cant see it well above.
anonymous
  • anonymous
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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