First integral:
Parameterize -C (the reverse of C) by [tex]z = e^{it}[/tex] where [tex]0\le t \le\frac\pi2[/tex]. Then the contour integral is
[tex]\displaystyle \int_{-C} (z^2 - z + 2) \, dz = \int_0^{\frac\pi2} \left(e^{2it} - e^{it} + 2\right) i e^{it} \, dt[/tex]
[tex]\displaystyle = i \int_0^{\frac\pi2} \left(e^{3it} - e^{2it} + 2e^{it}\left) \, dt[/tex]
[tex]\displaystyle = i \left(\frac1{3i} e^{3it} - \frac1{2i} e^{2it} + \frac2i e^{it}\right) \bigg|_0^{\frac\pi2}[/tex]
[tex]\displaystyle = \left(\frac13 e^{3it} - \frac12 e^{2it} + 2 e^{it}\right) \bigg|_0^{\frac\pi2}[/tex]
[tex]\displaystyle = -\frac43 + \frac53i[/tex]
Then
[tex]\displaystyle \int_C (z^2-z+2) \, dz = -\int_{-C} (z^2+z-2) \, dz = \boxed{\frac43 - \frac53i}[/tex]
Second integral:
The integrand has a pole of order 2 at z = 3i which is contained in C. By the residue theorem,
[tex]\displaystyle \int_C \frac{z^2+z}{(z-3i)^2} \, dz = 2\pi i\, \mathrm{Res}\left(\frac{z^2+z}{(z-3i)^2}, z=3i\right)[/tex]
Compute the residue:
[tex]\displaystyle \mathrm{Res}\left(\frac{z^2+z}{(z-3i)^2},z=3i\right) = \lim_{z\to3i} \frac{d}{dz}\left[z^2+z\right] = \lim_{z\to3i} (2z+1) = 1 + 6i[/tex]
Then the value of the integral is
[tex]\displaystyle \int_C \frac{z^2+z}{(z-3i)^2} \, dz = 2\pi i (1 + 6i) = \boxed{-12\pi + 2\pi i}[/tex]
