Asumming I have the following integral to solve in the complex plane: $$\int \frac{dz}{z+1} $$ while $|z|=5$ which means a contour of radius 5 around zero. Is it possible to solve this integral using: $$z+1=5e^{it}$$ while $t$ runs from $0$ to $2\pi$ ?

if this move is legit, what excactly am I doing? is it a parameterization or is it a varbile substitution?


That doesn't seem right.

$$z+1=5e^{it}$$ isn't a correct parameterization of the circle with radius 5 and the center as origin.

A correct parameterization would be: $z=z(t) = 5e^{it}$ which would lead to the integral:

$$\int_0^{2\pi}\frac{1}{5e^{it}+1}\cdot ie^{it} \text{d}t$$

Is it a variable substitution?

No, then you would need to apply the 'transformation of contour integration'-theorem:

If $w$ holomorphic on $\Gamma$ and $f$ continous on $w(\Gamma)$ then $$\int\limits_{w(\Gamma)} f(w) \operatorname dw = \int\limits_\Gamma f(w(z)) w'(z) \operatorname dz$$

Which would imply that you need to change the contour itself when transforming $z+1=5e^{it}$

How do I solve it?

There are various ways of solving this. You could use the residue theorem (which seems a bit overkill here).

Another possibility is to use the transformation of contour integrals and transforming $z +1 =w$, where the contour would shift 1 to the left.

Since the new contour still encloses the origin, you probably know the result: $2\pi i$

  • $\begingroup$ If I make a transformation like you suggested for $z+1=w$ what would my curve shift into? $\endgroup$ – user3921 Jan 10 '15 at 12:36
  • 1
    $\begingroup$ A circle with center $-1$ and radius 5. $\endgroup$ – dietervdf Jan 10 '15 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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