Evaluate $\int_{|z|=1} z^2 d\mu$

Question

Let $\mathbb{T} = \{z\in \mathbb{C}: |z| = 1\}$, $\phi:[0,1] \to \mathbb{T}$ be defined by $\phi(t) = e^{2 \pi i t}$, $m$ be the Lebesgue measure, and let $\mu(E) = m(\phi^{-1}(E))$. Find $$\int_{\mathbb{T}} z^2 d\mu$$

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I think that I should do the following: $$\int_{\mathbb{T}}z^2 d(m \circ \phi^{-1}) = \int_{\phi^{-1}(\mathbb{T})} (\phi(t))^2 dm = \int_0^1 e^{4\pi i t} dm = 0$$ But I'm worried I've oversimplified something here. Is this all I need to do?

Answer 1

That is all you need.

Alternatively, note that $d\mu$ is invariant under the rotation $z\mapsto iz$, so that $$\int_T z^2\,d\mu=\int_T (iz)^2\,d\mu$$ etc.