The theorem (from Wiki):
#\oint _{\gamma }f(z)\,dz=0#
The contour integral around the closed path should be zero.
I am going to use vector notation for the linear bits as its clearer and the algebra is the same
#vec{AB} = ((2),(2))#
So the param is # z = ((1),(1)) + t((1),(1)), t in [0,2]#
# dz = ((1),(1)) dt#
#implies int_(t = 0)^2 ( ((1),(1)) + t((1),(1)) + ((-1),(1))) ((1),(1)) dt #
#= ((1),(1)) int_(0)^2 ((0),(2)) + t((1),(1)) dt #
#= ((1),(1)) [ ((0),(2))t + t^2/2((1),(1)) ]_(0)^2 #
#= ((1),(1)) [ ((0),(4)) + ((2),(2)) ]#
#= ((1),(1)) ((2),(6))#
#= (1+i) ( 2+6i) #
#= 2 + 6i + 2i - 6#
#= -4 + 8i #
#vec{BC} = ((-4),(0))#
# z = ((3),(3)) + t((-1),(0)), t in [0,4]#
# dz = ((-1),(0)) dt#
#implies int_(0)^4 ( ((3),(3)) + t((-1),(0)) + ((-1),(1))) ((-1),(0)) dt #
#= ((-1),(0)) int_0^4 ((2),(4)) + t((-1),(0)) dt #
#= ((-1),(0)) [ ((2),(4))t + t^2/2((-1),(0)) ]_0^4 #
#= ((-1),(0)) ( ((8),(16)) + 8((-1),(0)) )#
#= ((-1),(0)) ((0),(16)) #
=#-1 * 16i#
#= -16i #
#vec{CA} = ((2),(-2))#
# z = ((-1),(3)) + t((1),(-1)), t in [0,2]#
# dz = ((1),(-1)) dt#
#implies int_0^2 ( ((-1),(3)) + t((1),(-1)) + ((-1),(1))) ((1),(-1)) dt #
#=((1),(-1))int_0^2 ((-2),(4)) + t((1),(-1)) dt #
#= ((1),(-1)) [ ((-2),(4))t + t^2/2((1),(-1)) ]_0^2 #
#= ((1),(-1)) ( ((-4),(8)) + ((2),(-2)) )#
#= ((1),(-1)) ((-2),(6)) #
#= (1-i)(-2 + 6i)#
#= -2 + 6i + 2i + 6#
#= 4 + 8i #
#implies int_(AB) + int_ (BC) + int_(CA) #
#((-4),(8)) + ((0),(-16)) + ((4),(8)) = ((0),(0))#
#= 0 #
