Please solve

Question 1 a) Solve the following integrals. 10 points 1) I1=C1f(z)dz 2) I1=C2f(z)dz where f(z)=3z2+11. C1 is the square (positive sense) whose vertices are located at (1,1),(1,1),(1,1),(1,1), and C2 is the circle given by z=4. You should get the same result. Can you explain why? Hint: you might use the formulae: 1+t2dt=tan1t+C,tan1z1tan1z1=tan1(1+z1z2z1z2)