- #1
Rectifier
Gold Member
- 313
- 4
The problem
I would like to solve:
$$ \bar{z} = z^n $$ where ##n## is a positive integer.
The attempt
## z = r e^{i \theta} \\ \\ \overline{ r e^{i \theta} } = r^n e^{i \theta n} \\ r e^{-i \theta} = r^n e^{i \theta n} ##
## r = r^n \Leftrightarrow true \ \ if \ \ n=1 \ \ or \ \ r=1##
## e^{-i \theta} = e^{i \theta n} \\ -\theta = \theta n + 2 \pi k, \ k = 0,1,2...n \\ -\theta - \theta n = 2 \pi k \\ -\theta(1+n) = 2 \pi k \\ \theta(n+1) = 2 \pi k_1, \ k_1 = 0,-1,-2,...,-n \\ \theta = \frac{2 \pi k_1}{(n+1)} ##
I don't understand the answer to this problem:
I am not sure how all real numbers could be a solution when n=1 and where they got ##z=0## and the condition ##n>1## from.
I would like to solve:
$$ \bar{z} = z^n $$ where ##n## is a positive integer.
The attempt
## z = r e^{i \theta} \\ \\ \overline{ r e^{i \theta} } = r^n e^{i \theta n} \\ r e^{-i \theta} = r^n e^{i \theta n} ##
## r = r^n \Leftrightarrow true \ \ if \ \ n=1 \ \ or \ \ r=1##
## e^{-i \theta} = e^{i \theta n} \\ -\theta = \theta n + 2 \pi k, \ k = 0,1,2...n \\ -\theta - \theta n = 2 \pi k \\ -\theta(1+n) = 2 \pi k \\ \theta(n+1) = 2 \pi k_1, \ k_1 = 0,-1,-2,...,-n \\ \theta = \frac{2 \pi k_1}{(n+1)} ##
I don't understand the answer to this problem:
If ##n=1##, then all real numbers are the solution.
If ##n>1##, then ##z=0## or ## z=e^{i\frac{2 \pi k_1}{(n+1)}} ## for ## \ k = 0,1,2...n ##
I am not sure how all real numbers could be a solution when n=1 and where they got ##z=0## and the condition ##n>1## from.