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

1. anonymous

$-i = e ^{-i \frac{\pi}{2}}$

2. anonymous

$z^5 = e ^{-i \frac{\pi}{2} +2\pi ki}$

3. anonymous

then factor out the i$z^5 = e ^{i \pi [ 2k-\frac{1}{2} ] }$

4. anonymous

take 5 root both sides$z= e ^{\frac{i \pi}{5} [2k -\frac{1}{2} ]}$

5. anonymous

then sub in k=-2,-1,0,1,2 to get the five solutions

6. anonymous

$e ^{\frac{-i \pi}{2} }$

7. anonymous

prob cant even even see that , ahh you can do the rest

8. anonymous

just subing in numbers

9. anonymous

thanks man, EE for life

10. anonymous

how did you know to use k=-2,-1,0,1,2?

11. anonymous

kinda guessing ( but not really ) you spead the values of k as even as possible between positive and negative numbers

12. anonymous

so if you had a power of 5 you would do as I did

13. anonymous

if you had power of 3 you use k= -1,0,1

14. 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

15. anonymous

oh ok, thanks :D

16. 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

17. anonymous

you have to keep going until you find n different solutions ( for the general eqn z^n = x+iy )

18. anonymous

thats why its easiet to have the values of k all following one another , and split over positive and negatives

19. anonymous

also, you should go through and make sure that the arguments of the solutions are in the range -pi<theta<= pi

20. anonymous

is $-i=e^\pi/2$ an identity or something?

21. anonymous

-i=e^(pi/2)

22. anonymous

-i = e^(-pi/2) , probably cant see it well above.

23. 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