Let $f$ and $\phi$ be continuously differentiable functions on the closed unit disc $D=\{\lambda\in \Bbb C; |\lambda|\leq1\}$ and suppose that there exists $u:[0,2\pi]\to[u_0, u_0+2\pi]$ continuously differentiable such that $$e^{iu(t)}=\phi(e^{it}), t\in[0,2\pi].$$ Let $$A=\int_0^{2\pi}|f(\phi(e^{it}))|^2dt.$$ The author of the book I am reading say that, taking te substitution $u=u(t)$ we conclude that
$$A=\int_{u_0}^{u_0+2\pi}|f(e^{iu})|^2\frac{1}{|\phi'(e^{it(u)})|}du.$$
I want to conclude this equality, but I didn't undestand why we should put a modulus on that factor... If we differenciate $e^{iu(t)}=\phi(e^{it})$ in $t$ we have $$ie^{iu(t)}\frac{du}{dt}=ie^{it}\phi'(e^{it}), t\in[0,2\pi].$$ If we could take modulus in this equality I could conclude what I want... but why could I do this? The formula for integration by substitution is $$\int_{\varphi(a)}^{\varphi(b)}f(x)dx=\int_a^bf(\varphi(t))\varphi'(t)dt.$$