<s>[INST] Groups with Complex Numbers I'm working on a quick proof related to the complex numbers.
Take $G$ to be the group of all nonzero complex numbers under
multiplication ($\circ$). Let $H$ be the set of of complex
numbers such that the sum of squares of the two real parts of the 
complex number is equal to 1. I want to show that $H$ under
multiplication is a subgroup of $G$. I can see how to do the
closure part of the proof, but I am having issues finding an
accurate inverse for each element, likely due to the fact that
I am uncertain about the identity element in the complex case.
 [/INST] Remember that the identity under multiplication is 1. So the inverse of $z \in \mathbb C$ is the reciprocal $\frac{1}{z}$. Since $|z| = 1$, we know $\left|\frac{1}{z}\right| = \frac{1}{1} = 1$.
We can compute this explicitly if you are not comfortable with the absolute value of a complex number. Let $z = x+iy$ such that $x^2+y^2=1$ and $x, y \in \mathbb R$. We now have $$\frac{1}{x+iy} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{x-iy}{x^2+y^2} = \frac{x}{x^2+y^2} + i\frac{-y}{x^2+y^2}.$$
Of course, we need to check that sum of squares of real and complex parts is 1. So we get
$$\left(\frac{x}{x^2+y^2}\right)^2 + \left(\frac{-y}{x^2+y^2}\right)^2= \frac{x^2+y^2}{(x^2+y^2)^2},$$ which is 1 because we assumed that $x^2+y^2=1$.
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