Proving that a subspace of C is a field

lolittaFarhat
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Homework Statement


Show that the only m for which the subspace of C given by {z ∈ C: Im(z) = m Re(z)} is a field is m=0.

Homework Equations


Field axioms

The Attempt at a Solution


I tried to prove one direction :
- If z is in the subspace, Re z>0 and m≠0 then Arg z<Arg z^2, so z^2 is not in the subspace.
I need a hint to prove the other direction.
 
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lolittaFarhat said:

Homework Statement


Show that the only m for which the subspace of C given by {z ∈ C: Im(z) = m Re(z)} is a field is m=0.

Homework Equations


Field axioms

The Attempt at a Solution


I tried to prove one direction :
- If z is in the subspace, Re z>0
No. For example, z = -1 -mi is in C, but Re(z) < 0. Note that for a given value of m, C is a straight line through the origin. Which field axioms prevent this subspace from being a field?
lolittaFarhat said:
and m≠0 then Arg z<Arg z^2, so z^2 is not in the subspace.
I need a hint to prove the other direction.
 
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An element of the subspace z would have the form z = x + imx, with x in the reals.
As Mark mentioned, you should check the axioms, it seems like this space satisfies most of the addition axioms. I think you were on the right track with squaring z. Using z^2 as an example, you can show that the only m that satisfies multiplication axioms (for any choice of x and a fixed m) is 0.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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