# Why does $|dz|=-ir\frac{dz}{z}$ when $|z|=r$?

Sometimes I want to compute a line integral over some circle $|z|=r$, where I have $|dz|$ instead of $dz$ given to me.

Reparametrizing with $z=re^{it}$, it follows that $dz=rie^{it}dt=izdt$. But I always read that $$|dz|=|iz|dt=|z|dt=|z|\frac{dz}{zi}$$ so $|dz|=-ir\frac{dz}{z}$. In the first equality, why does $|dz|=|iz|dt$ instead of $|iz||dt|$? Why doesn't the absolute value extend to the $dt$ as well?

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The absolute value extends to the $dt$ ($=d\phi$) as well; but when you integrate with respect to increasing $t$ then $|dt|=dt$. You could also compute the circumference of the circle by integrating "the wrong way around", but then you would have to take care of $|dt|=-dt$. – Christian Blatter Mar 19 '12 at 15:59

If a line integral is stated as $|dz|$, then it should be understood as integral of a function with respect to a measure (as opposed to the integral of a $1$-form). Such an integral does not depend on orientation. Therefore, it would be more correct to use $|dt|$ in place of $dt$ in the computation you quoted. The authors probably meant for $t$ to be increasing, in which case $|dt|$ and $dt$ are the same thing.