# Show that $\mathbb{T}^1 \cong \mathbb{C}^*/\mathbb{R}_{>0}$ [duplicate]

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I'm trying to use the First Isomorphism Theorem to show that $\mathbb{T}^1 \cong \mathbb{C}^*/\mathbb{R}_{>0}$ by constructing a surjective group homomorphism from the nonzero complex numbers to the circle group whose kernel is the set of positive reals. I haven't yet taken a course in complex variables, so I did some digging around and could only find a homomorphism from $\mathbb{R}^*$ to $\mathbb{T}^1$ defined by $f(\theta) = e^{i\theta}$. I don't really know what that means, so I'm struggling to construct any homomorphism $\phi : \mathbb{C}^* \to \mathbb{T}^1$, let alone one whose kernel is $\mathbb{R}_{>0}$.

EDIT: Does $f(z) = \frac{z}{|z|}$ work? EDIT2: Probably not. I think the image of this is actually just $\{\pm 1\}$. I really have no idea how to work with complex numbers at all.

## marked as duplicate by André 3000, Community♦Dec 12 '16 at 7:38

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• Dang, how did you find that? (in reference to it being marked as a duplicate) – playitright Dec 12 '16 at 7:37

## 2 Answers

Your homomorphism suffices as we have $$z=re^{i\phi}$$ and as such we get $$f(z)=\frac{re^{i\phi}}{|re^{i\phi}|}=\frac{re^{i\phi}}{r}=e^{i\phi}$$ which is the unit circle, and it is surjective.

• Thank you! But, how do you know that $ker(f) = \mathbb{R}_{>0}$? – playitright Dec 12 '16 at 7:06
• Because the multiplicative identity is $1$ with $\phi=0$ and as such all real numbers are sent to $1$ by that homomorphism. – Zelos Malum Dec 12 '16 at 7:07

The homomorphism you obtained works: $$z\in\ker f\iff z/|z|=1\iff z=|z|\iff z\in\mathbb{R}_{>0}$$ The idea is that a complex number is given by an angle/argument, or equivalently an element of the unit circle, and its length/absolute value (this is the polar form). In fact, this is the argument to show that $f$ is surjective.