# Thread: Field C - linear dependence

1. ## Field C - linear dependence

Hello everyone, I'm new and need some help

I've been struggling two questions for like two hours already. Sorry in advance if this sounds unclear, english isn't my native language and I'm not sure of the english names for what I refer to -

1) I have a collection of complex numbers vectors in C^3.
I need to find if they're independent when C^3 is linear map over C, and same if C^3 is linear map over R.
What is the difference in dependency if C^3 is over C or over R? The question is not if they're a linear map so the field, be it C or R, has no meaning, no? :S I don't understand it.

2) Need to solve the following equation:
z^4 = (-\sqrt(3) + i ) / ( -1 - i * \sqrt(3) )

I'll start by saying that I spend too much time doing it all over and over again. I'm not even sure how is this supposed to look :S
Is it
z^4 = ( \sqrt(3)*i^2 + 4i + \sqrt(3) )/4 ? :S

If it is, how do I go from here? What do I do?

Thanks in advance!

2. Originally Posted by Rita.g
Hello everyone, I'm new and need some help

I've been struggling two questions for like two hours already. Sorry in advance if this sounds unclear, english isn't my native language and I'm not sure of the english names for what I refer to -

1) I have a collection of complex numbers vectors in C^3.
I need to find if they're independent when C^3 is linear map over C, and same if C^3 is linear map over R.
"Linear map" isn't the correct word here. You mean "Linear vector space".

What is the difference in dependency if C^3 is over C or over R? The question is not if they're a linear map so the field, be it C or R, has no meaning, no? :S I don't understand it.
A set of vectors is "dependent" over a given field if there exist numbers in the field, $a_1$, $a_2$, etc., not all 0, such that $a_1v_1+ a_2v_2+ \cdot\codt\cdot+ a_nv_n= 0$. The difference between "over C" and "over R" is whether the numbers are allowed to be complex or real. In particular, C^3 "over C" has dimension 3 while "over R" it has dimension 6. Do you see why?

2) Need to solve the following equation:
z^4 = (-\sqrt(3) + i ) / ( -1 - i * \sqrt(3) )

I'll start by saying that I spend too much time doing it all over and over again. I'm not even sure how is this supposed to look :S
Is it
z^4 = ( \sqrt(3)*i^2 + 4i + \sqrt(3) )/4 ? :S
You are aware that $i^2= -1$, surely! That would be just i. But you are missing a sign. It should be (-\sqrt(3)+ 3i- i+ \sqrt{3})/4= 2i/4= i/2.

If it is, how do I go from here? What do I do?

Thanks in advance!
Now write that in "polar form" (which is very easy for i/2) and use "DeMoivres' formula". Writing $z= re^{i\theta}= r(cos(\theta)+ i sin(\theta))$, the nth root is given by $z^{1/n}= r^{1/n}e^{\frac{\theta+ 2\pi k}{n}= r^{1/n}(cos(\frac{\theta+ 2\pi k}{n}+ i sin(\frac{\theta+ 2\pi k}{n})$ where $r^{1/n}$ is the positive nth root of the positive real number r and you can get all n roots by taking k= 0, 1, 2, 3, ..., n-1.

3. Thank you for the quick reply!

Originally Posted by HallsofIvy
The difference between "over C" and "over R" is whether the numbers are allowed to be complex or real. In particular, C^3 "over C" has dimension 3 while "over R" it has dimension 6. Do you see why?
But if I found that the vectors are indeed dependent, can their dependency be changed based on whether they're over C or R?
As for dimension 6 - I'm clueless...

Originally Posted by HallsofIvy
You are aware that $i^2= -1$, surely! That would be just i. But you are missing a sign. It should be (-\sqrt(3)+ 3i- i+ \sqrt{3})/4= 2i/4= i/2.
Ohh right, i^2! Silly me, overlooked that. Thank you!
But I don't understand how is the first sqrt(3) is negative, it is multiplied by -sqrt(3) and -1?