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n is a natural number

I am only having a problem with $r=0.3$

It can be written as $$\cos (0.3\pi n) + i\sin (0.3\pi n)$$

Now clearly the expression varies with the value of $n$ so it should be infinite, but it is actually finite. What am I missing?

Aditya
  • 6,191
  • It's because $r=0.3$ is rational, have a look at this. – rtybase Jan 27 '21 at 18:33
  • Note that $20\pi r = 6\pi$ – Paul Sinclair Jan 27 '21 at 20:34
  • @rtybase I am not aware about the concept of density – Aditya Jan 28 '21 at 01:27
  • @PaulSinclair can you elaborate? I am not sure what I am supposed to do with that info – Aditya Jan 28 '21 at 01:28
  • Did you even look? Did you actually ask yourself what $e^{6\pi i}$ is? – Paul Sinclair Jan 28 '21 at 01:34
  • @PaulSinclair I did look, the value is 1. What does that have anything to do with $0.3\pi$ – Aditya Jan 28 '21 at 01:36
  • Note that ${\bf 20\pi r} = 6\pi$ – Paul Sinclair Jan 28 '21 at 01:46
  • @PaulSinclair I am pretty sure it is really easy and it’s right in front of me but I am just not able to identify it. I get that $20\pi r=6\pi$. How does that help – Aditya Jan 28 '21 at 03:27
  • You are talking about the set $A_r ={e^{i\pi r,n}\mid n \in \Bbb N}$. When $r=0.3, n = 20$, the value of $e^{i\pi r,n} = 1$. Now figure out what that means for other values of $n$. For example, can your think of any other values of $n$ that yield $1$? Is there is some property of exponentiation that might be useful here? Come on. You cannot have gotten this far without knowing something about how these functions work. Make use of that information. – Paul Sinclair Jan 28 '21 at 17:26
  • @PaulSinclair here is what I understood from your comment. You either meant $(e^{0.3\pi i})^{20}=1 \implies e^{0.3 \pi i}=1$, which seems a bit odd (that’s just my interpretation) or you are talking about other values of $n$ which gives the same result, which are infinite. $n=18,60, ....$. So, what am I missing here? I apologize if I am being dumb here, but I genuinely cannot understand what you are try to say. Yes, I do know how these functions work, but I am just not able to apply it right now. – Aditya Jan 29 '21 at 01:48
  • Okay - I can't blame you for running afoul of that, but no, $(e^{0.3\pi i})^20 = 1$ does not imply $e^{0.3\pi i} = 1$. Those laws of exponents you are familiar with mostly only hold when the base is a positive integer. When the base is negative or complex they break down except for integer exponents. But the only base you need to figure this out is $e$, for which all the rules hold. The rule I was hinting at $e^{a+b} = e^a + e^b$. Or you can use D'Moivre's thjeorem on $\cos(nr\pi) + i\sin(nr\pi)$. $n=18$ do not give $1$, but $n=0$ does. As does $n=40, 60, 80, ...$. – Paul Sinclair Jan 29 '21 at 04:30
  • I'll give you one more hint: $A_{0.3}$ has exactly 20 elements. And the reason is $20(0.3)\pi i = 6\pi i$. – Paul Sinclair Jan 29 '21 at 04:31
  • @PaulSinclair I didn’t know $e^{a+b}=e^a+e^b$ was a thing – Aditya Feb 01 '21 at 18:14
  • That was typo, as anyone who was trying to be a smartass would recognize. If you don't know the actual formula, then you should be taking remedial algebra, not this course. – Paul Sinclair Feb 01 '21 at 19:01
  • @PaulSinclair I wasn’t trying to be a smartass, I genuinely thought that it was some other method of solving complex numbers. – Aditya Feb 02 '21 at 01:21
  • My apologies then! No I meant $e^(a+b) = e^ae^b$. In general, if you see $z^w$, it is only well-defined if $z$ is a positive real number, or $w$ is an integer. Even $(-1)^{1/2}$ is indeterminant between $i$ and $-i$. If $z$ is complex or negative, and if $w$ is not integer (including being complex), then if you need to use $z^w$ you have to be clear as to which of the many possible values of the expression you mean. However, when $z > 0$ or $w$ is integer, then the expression is well-defined and follows the exact same rules as you learned. – Paul Sinclair Feb 02 '21 at 02:29

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