Can complex numbers be used to show a pattern in exponents?

[icystrike]

Homework Statement

Given that $$(1-\sqrt{3}i)^{n}$$ is real and positive , use de Moivre's Theorem to show that the values of n are terms of arithmetic progression.

Homework Equations

The Attempt at a Solution

I've worked it out . n=3k s.t k element of positive integers.

 

[Mark44]

Are you sure you have given us the problem exactly as stated? My guess is that they're asking you to show that the values of (1 - sqrt(3)i)ⁿ are terms of an arithmetic progression. As you've stated the problem it doesn't make much sense.

 

[icystrike]

Are you sure you have given us the problem exactly as stated? My guess is that they're asking you to show that the values of (1 - sqrt(3)i)ⁿ are terms of an arithmetic progression. As you've stated the problem it doesn't make much sense.

Hi Mark! I've checked with the question again , and its exactly what I've stated.

 

[Mark44]

OK - just checking. Apparently the problem asks you to find which values of n are such that (1 - sqrt(3)i)^n are real and positive. What is 1 - sqrt(3)i in polar form? That should help you figure out a lot more easily what the powers of 1 - sqrt(3)i look like.

 

[icystrike]

OK - just checking. Apparently the problem asks you to find which values of n are such that (1 - sqrt(3)i)^n are real and positive. What is 1 - sqrt(3)i in polar form? That should help you figure out a lot more easily what the powers of 1 - sqrt(3)i look like.

yeap. I've worked it out in my attempted solution and I'm convinced that it is a arithmetic progression as n has 0 as first term and 3 as common difference for each term.

 

[Mentallic]

Not quite.

The question is asking for when it's real and positive. Since 2ⁿ is always positive for natural n, then you are searching for when the complex part has an argument of 0. i.e. $\theta=2k\pi$ where k is all integers.

You can even see for yourself that for n=3 the value of $$\left(1-\sqrt{3}i\right)^3$$ is real, but negative.

 

[HallsofIvy]

yeap. I've worked it out in my attempted solution and I'm convinced that it is a arithmetic progression as n has 0 as first term and 3 as common difference for each term.

You still don't understand what Mark44 was asking.

In your original post, you gave a complex number to the "n" power and asked to "show that the values of n are terms of arithmetic progression."

Well, that's trivial! The values of n are 1, 2, 3, 4, 5, ..., with common difference 1!

 

[Mark44]

You still don't understand what Mark44 was asking.

In your original post, you gave a complex number to the "n" power and asked to "show that the values of n are terms of arithmetic progression."

Well, that's trivial! The values of n are 1, 2, 3, 4, 5, ..., with common difference 1!

1, 2, 3, ... don't work in the problem as I understand it. IMO the problem is poorly stated and misleading. A better restatement of the problem might be "For what n is the expression (1 - sqrt(3)i)ⁿ real and positive? Show that these values of n form an arithmetic progression."

 

[Mentallic]

I believe the question was clear enough. It first mentions that $$(1-\sqrt{3})^n$$ is real and positive, which obviously indicates that n is not all natural numbers. Then it goes on to say show that the values of n are in arithmetic progression.
Shouldn't be confusing whatsoever in my opinion.

 

[Mark44]

I still maintain that the problem could have been clearer if rephrased. "Given that (1 - i*sqrt(3))^n is real and positive, ..." would cause one to reasonably infer that this statement was true for integer values of n, or possibly just nonnegative integer values.

 

[Mentallic]

would cause one to reasonably infer that this statement was true for integer values of n

Maybe for someone without a basic knowledge of complex numbers and exponentials, but this is untrue for everyone that has posted in this thread.

Also, I believe that realizing the base case of natural numbers n=1 contradicts the statement that $(1-i\sqrt{3})^n$ is real and positive would be easier than having to make some reasonable judgement on what the question is saying.

... I stand by my statement that it's obvious what the question was asking.

 

[HallsofIvy]

Ah, thanks to both Mark44 and Mentallic. I completely misunderstood the problem. It might be better phrased "show that those n such that $(1- i\sqrt{3})^n$ are real and positive form an arithmetic progression".

 

[Mark44]

Maybe for someone without a basic knowledge of complex numbers and exponentials, but this is untrue for everyone that has posted in this thread.

Also, I believe that realizing the base case of natural numbers n=1 contradicts the statement that $(1-i\sqrt{3})^n$ is real and positive would be easier than having to make some reasonable judgement on what the question is saying.

... I stand by my statement that it's obvious what the question was asking.

Ah, thanks to both Mark44 and Mentallic. I completely misunderstood the problem. It might be better phrased "show that those n such that $(1- i\sqrt{3})^n$ are real and positive form an arithmetic progression".

I rest my case.

 

[Mentallic]

Ah, thanks to both Mark44 and Mentallic. I completely misunderstood the problem. It might be better phrased "show that those n such that $(1- i\sqrt{3})^n$ are real and positive form an arithmetic progression".

would cause one to reasonably infer that this statement was true for integer values of n

Maybe for someone without a basic knowledge of complex numbers and exponentials, but this is untrue for everyone that has posted in this thread.

I guess I was wrong...

Haha just kidding