Thanks for any help in advance.
I'm working on question 11 in the following image
The question is as follows:
Evaluate the contour integral (closed loop)
$$ \int z^*-3 $$
over the path anticlockwise (of which there are three parts):
- (0,0) to (2,0) (the line y = 0)
- (2,0) to (0,2i) over the line $\mod{z}$ = 2 (a circle of radius 2 centred on the origin in the first quadrant)
- (0,2i) to (0,0) (the over the line z = 0)
My initial thoughts are that the $z^*$ function does not satisfy the Cauchy Riemann conditions, and so it is not complex differentiable. Hence we can't use Cauchy's Theorem to say that the closed loop integral is equal to zero. But that shouldn't stop it from being integrable, just that it returns different values for different paths from the same endpoints.
Secondly, there are no isolated singularities, so i can't use the second Cauchy's integral theorem (and even if there were any the region isn't analytical anyway so i wouldnt be able to apply it)
I evaluated the integral over the three segments, getting:
- $-4$
- $2i\pi -6i+6$ (using the substitution $z^*=4/z$)
- $-4$
However, my answer differs from the actual answer ($2\pi i$)
Is my reasoning, or my calculation (or both) wrong?
From $\;(0,0)\;$ to $\;(2,0)\;$ , straight line: $\;t(2,0)+(1-t)(0,0)=(2t,0)\;,\;\;0\le t\le 1:\;$
$$\int_c\overline z-3=\int_0^1\left((t-0i)-3\right)\cdot2\,dt=\int_0^1(2t-6)\,dt=1-6=\color{red}{-5}$$
From $\;(2,0)\;$ to $\;(0,2)\sim 2i\;$ over
$$\;r(t):=(2\cos t,\,2\sin t)\sim 2(\cos t+i\sin t)=2e^{it}\;,\;r'(t)=2ie^{it},\,\;0\le t\le \frac\pi2:$$
$$\overline z-3=e^{-it}-3\implies $$
$$\int_c\overline z-3 =\int_0^{\pi/2}\left(2e^{-it}-3\right)2ie^{it}\,dt=\int_0^{\pi/2}\left(4i-6ie^{it}\right)\,dt=$$
$$=2\pi i-\left.6i\frac1ie^{it}\right|_0^{\pi/2}=2\pi i-6(e^{\pi i/2}-e^0)=\color{red}{2\pi i-6i+6}$$
From $\;(0,2)\sim 2i\;$ to $\;(0,0)\;$ :$\;(0,(2(1-t))\;,\;\;0\le t\le 1\;$ :
$$\int_c\overline z-3=\int_0^1\left(0-it-3\right)(-2i\,dt)=\int_0^1\left(-2t+6i\right)\,dt=\color{red}{-1+6i}$$
Sum all and you get $\;2\pi i\;$ ...