Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.