- #1

- 28

- 0

_{C}(z)/z

^{2}+9 dz , where C is the circle │z-2i│=4.

what i have done so far is :

z(t) = 2i + 4e

^{it}

z'(t) = 4ie

^{it}

f(z(t)) = 4ie

^{it}/(4ie

^{it})

^{2}+9

∫ (4ie

^{it}/(4ie

^{it})

^{2}+9) (4ie

^{it}) dt

intergrate from 0->2pi

but i dont know how to solve this intergral, can anyone help?

2. ∫

_{c}cos(z)/(z-1)^3(z-5)^2 dz , where C is the circle │z-4│=2.

this z'(t) = 0

so , is this intergral equal 0?

since f(z(t))(z'(t)) = 0

Thanks