The theorem (from Wiki):
\oint _{\gamma }f(z)\,dz=0∮γf(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
