Rectifier said:
I understand the part with k but, unfortunately, I still don't get the rest of it, sorry. I suspect that I misunderstand some very basic algebra.
Let me try again and provide a few more steps to my calculation:
## r=r^n \\ r \left(1-r^{n-1} \right) = 0##
This is the correct way to look at it, yes.
I understand this as ## r_1 = 0 ## if ## n \neq 1 ##. Since if ##n = 1## then we have ## r \cdot \left(1-r^{1-1} \right) \Rightarrow r \cdot 0 = 0 ## which is solved by all ##r##:s.
## 1-r^{n-1} = 0 ## leads to ## 1=r^{n-1} ## which leads to ## r_2 = 1^{\frac{1}{n-1}} ## therefore ## r_2 = 1 ##.
Yes. You could also just say: ##r=0 \,\, \text{ OR } \,\, 1-r^{n-1}=0##.
This way you have the logic or in between, which means, and is allowed. They can both be zero, which answers your previous question in post #3. The distinction in the book starts with ##n##. ##n## is either ##1## or greater than ##1##. This separates all further consideration into these two parts. As the result is a combination of ##n,r,\theta##, the combinations will be distinguished by ##n##, not by the other two numbers. There can and will still be ##(1,0,\theta)## and ##(n>1,0,\theta)##, which are different as triples, although resulting in the same number for ##z##. Similar happens to ##\theta## where you have many values resulting in the same number ##e^{i \theta}##. However, they are still different as triples, because we separated the cases ##n=1## and ##n>1##. I hope this clarifies your statement:
Since ## z=re^{i\theta n} ## then ## r_1 = 0 ## leads to ## z=0 \cdot e^{i\theta n} \Rightarrow z = 0## when ## n \neq 1 ##. We now that ## n \in \mathbb{Z^+} ## "without 1" - ## n \in \mathbb{Z^+} \setminus \{1\} ## which is bascially ## n>1 ##. It is bascially the same thing for z where ## r_2 = 1 ## and we'll get the same result as in my first post. This part I understand, somewhat. Now to the next one.
further.
Now for ##n=1##. How do I reason here?
I guess I could go back to ## r=r^n \\ r \left(1-r^{n-1} \right) = 0## and set ## n =1 ## and get ## r \cdot 0 = 0 ##. All ##r## solve it but I am expecting to get ## z=r## from ## z=re^{i\theta n} ## . In that case we would need ## \theta = 0 ## since ## n = 1 ##, ## z=re^{i 0 \cdot 1} ##. But how could one reason here?
You have forgotten the step before ##r \left(1-r^{n-1} \right) = 0##.
The reason why we got this equation is, that we looked at ##\overline{re^{i \theta}} = r^ne^{i n \theta}## and compared only their length:
$$
\overline{re^{i \theta}} = r^ne^{i n \theta} \Rightarrow |\overline{re^{i \theta}}| = |r^ne^{i n \theta}| \Rightarrow r = r^n \Rightarrow r(1-r^{n-1}=0 \Rightarrow r=0 \,\, \text{ OR }\,\, n=1
$$
That means we deduced a necessary condition which must hold, if ##\bar{z}=z^n##. It means the absolute value (length) of our complex numbers ##\bar{z},z^n## has to be equal. It does not mean, that automatically all such numbers are a solution. So ##r=0 \,\, \text{ OR } \,\, n=1## has always to hold. Now we can distinguish the cases ##n=1## and ##n>1##. One of them has to be true, and only one.
CASE 1: ##n=1##
This means ##\bar{z}=z^1=z## which is equivalent to ##z \in \mathbb{R}## which automatically means ##\theta = 0## per convention.
Now we formally will have to check, whether all reals are actually a solution. O.k. this is easy and true. Thus all ##(n,r,\theta) = (1,r,0)## are a solution, which still means ##z## is real.
CASE 2: ##n>1##
This means ##r=0 \Leftrightarrow z=0 \,\, \text{ OR } \,\, r=1## are the only solutions. The case ##r=0## is clear. Now from here we have to consider more cases:
Given ##n>1, r=1##, which are necessary condition on ##\theta\,?## You have found that under these circumstances, ##\theta = \dfrac{2 \pi k}{n+1}## for all ##k \in \mathbb{Z}##. This is still a necessary condition, because we only have considered conclusions from ##\bar{z}=z^n##.
Now we formally will have to check, whether all those complex numbers ##z=1 \cdot e^{i \theta}##
are actually a solution. This is the case, so all possibilities are checked.
In sum we have
$$
\{z=re^{i\theta}\triangleq (n,r,\theta)\,\vert \,\bar{z}=z^n\} = \{(n,r,\theta)\,\vert \,n=1 \wedge r\in \mathbb{R}\wedge \theta \stackrel{(*)}{=} 0\} \cup \{(n,r,\theta)\,\vert \, n>1 \wedge r=0 \wedge \theta \stackrel{(*)}{=} 0 ) \} \cup \{(n,r,\theta)\,\vert \, n>1 \wedge r=1 \wedge \theta =\frac{2 \pi k}{n+1} (k \in \mathbb{Z}) \}
$$
and the unions are not all disjoint and ##(*)## per convention of how the reals are considered as complex numbers.
By the way, the formal step from ##1 \cdot e^{-i \frac{2 \pi k}{n+1}}= 1 \cdot e^{i \frac{2 \pi k n}{n+1}}## is to multiply the equation by ##e^{+ i \frac{2 \pi k}{n+1}}## and investigate when the new exponent will be zero to match the ##1=e^0## on the left hand side.