The arclength for a continuously differentiable curve is given by: $$\int|\gamma'(t)|dt=\int\gamma'(t)\overline{\gamma}'(t)dt$$
Consider $\gamma(t)=\exp(2it),t\in[0,2\pi]$
Then arclength is computed: $$\int_0^{2\pi}2i\exp(2it)*-2i\exp(-2it)dt = 8\pi$$
But $\exp(2it)=\cos(2t)+i\sin(2t)$ so I expect to trace around the unit circle twice if $t\in[0,2\pi]$ -- giving an arclength $4\pi$.
I suspect I'm missing something obvious, but not sure what though.
As mentioned in the comments, the issue is the lack of a square root in the first displayed equation.