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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[i = e ^{i \frac{\pi}{2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ z^5 = e ^{i \frac{\pi}{2} +2\pi ki}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then factor out the i\[z^5 = e ^{i \pi [ 2k\frac{1}{2} ] } \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take 5 root both sides\[ z= e ^{\frac{i \pi}{5} [2k \frac{1}{2} ]}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then sub in k=2,1,0,1,2 to get the five solutions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[e ^{\frac{i \pi}{2} }\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0prob cant even even see that , ahh you can do the rest

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just subing in numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks man, EE for life

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you know to use k=2,1,0,1,2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0kinda guessing ( but not really ) you spead the values of k as even as possible between positive and negative numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if you had a power of 5 you would do as I did

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you had power of 3 you use k= 1,0,1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0nevertheless , 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
 5 years ago
Best ResponseYou've already chosen the best response.0you have to keep going until you find n different solutions ( for the general eqn z^n = x+iy )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats why its easiet to have the values of k all following one another , and split over positive and negatives

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also, you should go through and make sure that the arguments of the solutions are in the range pi<theta<= pi

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is \[i=e^\pi/2\] an identity or something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i = e^(pi/2) , probably cant see it well above.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol 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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