Verifying $\mathbb T$ a subgroup of the multiplicative group of non-zero complex numbers

Question

Question: Let $\mathbb T$={$z\in$ $\mathbb Z$ :$\vert z\vert$=$1$}.
Verify $\mathbb T$ is a subgroup of the multiplicative group of non-zero complex numbers?

I do not know what this type of $\mathbb T$ actually means so I am not understanding the problem. I have never encountered a question like this. Need some help

Answer 1

I'm assuming that that should be $\mathbb{C}$ and not $\mathbb{Z}$. If so the given $z \in \mathbb{C}$, then $|z| = |a + bi| = \sqrt{a^{2} + b^{2}}$. Thus the question is asking you to show that such a set with $|z| = 1 \implies \sqrt{a^{2} + b^{2}} = 1 \implies a^{2} + b^{2} = 1$ is a subgroup of $\mathbb{C}^{*}$ under multiplication. Thus you must show that the identity is in $\mathbb{T}$, that for all $a, b \in \mathbb{T}, ab \in \mathbb{T}$ and for all $a \in \mathbb{T}$ $ \exists a^{-1} \in \mathbb{T}$ such that $aa^{-1} = 1$.