64.4k views
3 votes
Find the integral along the counterclockwise curve C:∣z∣=1 ∫​ i/zsin2(z)​dz. a.0 b.1 c.2πi d.πi/6​ e.−πi/4​ f.πi g.iπ/2​ h.2πi/3​

User DChhapgar
by
7.9k points

1 Answer

3 votes

Final answer:

The integral along the counterclockwise curve C: |z|=1 ∫​ i/zsin2(z)​dz is 0.

Step-by-step explanation:

To find the integral along the counterclockwise curve C: |z|=1 ∫​ i/zsin2(z)​dz, we can use the Residue Theorem. Since the function has a singularity at z=0, we need to find the residue at that point. The residue can be found by taking the limit as z approaches 0 of (z-0) * sin2(z).

By factoring out z from the numerator

Using L'Hopital's Rule, we can differentiate the numerator and denominator, resulting in z * (2z*sin(z)*cos(z) + z2*sin2(z))/(2z*sin(z)*cos(z)).

The term (2z*sin(z)*cos(z))/(2z*sin(z)*cos(z)) simplifies to 1, so the limit as z approaches 0 is z^2.

Therefore, the residue at z=0 is 02 = 0.

Since the curve C: |z|=1 encloses the singularity at z=0, we can use the Residue Theorem to find the value of the integral. The Residue Theorem states that the integral along a closed curve of a function with a singularity is equal to 2πi times the sum of the residues of the singularities enclosed by the curve.

Since the residue at z=0 is 0, the integral along the curve C: |z|=1 ∫​ i/zsin2(z)​dz is equal to 2πi times 0, which is 0.

User John Washam
by
8.4k points

Related questions

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.