Question

how can I explain using z^n=r^n(cos(ntheta)+isin(theta)) that z^1/n=r^1/n(cos((ntheta+2kpi)/n)+isin((theta+2kpi)/n)))? and that k= 0,1,2,...,n-1 (so we get n solutions out)

Answer 1

whoah we are going to need to clean this up a little

Answer 2

\[z^n = r^{n} \cos(n \theta)+i \sin(\theta)\]
\[\frac{ z^1 }{ n } = \frac{ r^1 }{ n }\cos(\frac{ n \theta + 2k \pi }{ n })+ i \sin(\frac{ \theta+2k \pi }{ n })\]

Answer 3

did i do that right?

Answer 4

yes! besides the ntheta after the cos. it should be just theta

Answer 5

for both equations?

Answer 6

no just the bottom. because it becomes 1/n instead of n

Answer 7

\[z^n = r^n(\cos(n \theta)+i \sin(\theta))\] but this is correct?

Answer 8

yep

Answer 9

wait no. there is n in the sin

Answer 10

\[\frac{ z^1 }{ n }=\frac{ r^1 }{ n }(\cos(\frac{ \theta + 2k \pi }{ n })+i \sin(\frac{ \theta +2k \pi }{ n }))\]

Answer 11

exactly

Answer 12

\[z^n=r^n(\cos(n \theta)+i \sin(n \theta))\]

Answer 13

okay

Answer 14

so then i need to use the top equation to show how you get bottom one (where its 1/n)

Answer 15

\[z^n=r^n(\cos(n \theta)+i \sin(n \theta))\]
\[\frac{ z^1 }{ n }=\frac{ r^1 }{ n }(\cos(\frac{ \theta + 2k \pi }{ n })+i \sin(\frac{ \theta +2k \pi }{ n }))\]
k = 0, 1, 2 ... n-1

Answer 16

im assuming it is an identity

Answer 17

it is used for taking nth root of a complex number

Answer 18

where z could be 1+3i?

Answer 19

yes

Answer 20

but how do I show that one leads to the other one? like where does the nkpi comes from? (i understand that it's adding another circle but how do I explain it in the equation? and why do I get n-1 solutions from it

Answer 21

first of all, you're adding 2kpi to the numearator inside angle, not nkpi

Answer 22

oops i meant 2kpi

Answer 23

lets start from De Moivre's theorem for integer powers :

\[\large z^n = r^{n} \cos(n \theta)+i \sin(n\theta)\]

Answer 24

is that what you want to start with ?

Answer 25

yes

Answer 26

if a function is periodic with a period of \(T\), can we say :
\[f(x) = f(x+T)\]

cuz its value repeats every \(T\) units of x

Answer 27

2pi isnt it?

Answer 28

yes period of \(\cos (\theta)\) is \(2\pi\)
so can we write \(\cos \theta = \cos (\theta + 2\pi)\) ?

Answer 29

and how about :
\[\cos \theta = \cos (\theta + 2\pi) = \cos(\theta +2\pi + 2\pi+2\pi + \cdots)\]

?

Answer 30

yea i get this part

Answer 31

instead of that messy looking expression, we say :
\[\cos(\theta) = \cos(\theta + 2k\pi)\]

where k = 0,1,2,3,...
(also negatives)

Answer 32

good, lets start with a complex number

Answer 33

\[z = \cos(\theta) + i\sin(\theta)\]

which is same as

\[z = \cos(\theta + 2k\pi) + i\sin(\theta + 2k\pi)\]

yes ?

Answer 34

because both sin and cos are periodic and ofcourse k is some integer

Answer 35

yes

Answer 36

*any integer

Answer 37

*all integers

Answer 38

good, next rise it to 1/n power using De moivre's thm

Answer 39

\[z = \cos(\theta) + i\sin(\theta)\]

which is same as

\[z = \cos(\theta + 2k\pi) + i\sin(\theta + 2k\pi)\]
By De Moivre's thm :

\[z = \cos(\theta) + i\sin(\theta)\]

which is same as

\[z^{1/n} = \cos\left(\dfrac{\theta + 2k\pi}{n}\right) + i\sin\left(\dfrac{\theta + 2k\pi}{n}\right)\]

Answer 40

k can be ANY integer,
however any "n" consecutive values of "k" gives u distinct solutions

Answer 41

"n" consecutive values of k could be :

k = 0,1,2,3,... n-1

or

k= 1,2,3,4,...n

or any other sequence of "n" consecutive integers will do

Answer 42

this is the part I don't understand
how come you get n solutions for every equation

Answer 43

thats a good question, lets see the solutions in complex plane

Answer 44

lets say if I had n=4, i will plug in k as 0,1,2,3 and get four solutions.

Answer 45

yes and those 4 solutions will be DISTINCT
meaning, they will be 4 different points in complex plane

Answer 46

lets take a number and take its nth root

Answer 47

pick some number

Answer 48

why