Question

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

Answer 1

\[-i = e ^{-i \frac{\pi}{2}}\]

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

then sub in k=-2,-1,0,1,2

to get the five solutions

Answer 6

\[e ^{\frac{-i \pi}{2} }\]

Answer 7

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

Answer 8

just subing in numbers

Answer 9

thanks man, EE for life

Answer 10

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

Answer 11

kinda guessing ( but not really )

you spead the values of k as even as possible between positive and negative numbers

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

oh ok, thanks :D

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

-i=e^(pi/2)

Answer 22

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

Answer 23

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