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.