Question

Please slove the problem in the below pic

Answer 1

First integral:

Parameterize -C (the reverse of C) by
`z = e^(it)` where
`0\le t \le\frac\pi2`. Then the contour integral is

`\displaystyle \int_(-C) (z^2 - z + 2) \, dz = \int_0^(\frac\pi2) \left(e^(2it) - e^(it) + 2\right) i e^(it) \, dt`

`\displaystyle = i \int_0^(\frac\pi2) \left(e^(3it) - e^(2it) + 2e^(it)\left) \, dt`

`\displaystyle = i \left(\frac1{3i} e^(3it) - \frac1{2i} e^(2it) + \frac2i e^(it)\right) \bigg|_0^(\frac\pi2)`

`\displaystyle = \left(\frac13 e^(3it) - \frac12 e^(2it) + 2 e^(it)\right) \bigg|_0^(\frac\pi2)`

`\displaystyle = -\frac43 + \frac53i`

Then

`\displaystyle \int_C (z^2-z+2) \, dz = -\int_(-C) (z^2+z-2) \, dz = \boxed{\frac43 - \frac53i}`

Second integral:

The integrand has a pole of order 2 at z = 3i which is contained in C. By the residue theorem,

`\displaystyle \int_C (z^2+z)/((z-3i)^2) \, dz = 2\pi i\, \mathrm{Res}\left((z^2+z)/((z-3i)^2), z=3i\right)`

Compute the residue:

`\displaystyle \mathrm{Res}\left((z^2+z)/((z-3i)^2),z=3i\right) = \lim_(z\to3i) (d)/(dz)\left[z^2+z\right] = \lim_(z\to3i) (2z+1) = 1 + 6i`

Then the value of the integral is

`\displaystyle \int_C (z^2+z)/((z-3i)^2) \, dz = 2\pi i (1 + 6i) = \boxed{-12\pi + 2\pi i}`