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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):

  1. (0,0) to (2,0) (the line y = 0)
  2. (2,0) to (0,2i) over the line $\mod{z}$ = 2 (a circle of radius 2 centred on the origin in the first quadrant)
  3. (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:

  1. $-4$
  2. $2i\pi -6i+6$ (using the substitution $z^*=4/z$)
  3. $-4$

However, my answer differs from the actual answer ($2\pi i$)

Is my reasoning, or my calculation (or both) wrong?

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  • $\begingroup$ I'm guessing that the second path is from the complex $\;2\sim (2,0)\;$ to the complex $\;2i\sim (0,2)\;$ . Am I right? I wrote the equivalences "complex $\,\leftrightarrow\,$ point on the plane $\;\Bbb R^2\;$ " $\endgroup$ – DonAntonio Apr 24 '18 at 11:28
  • $\begingroup$ yep, that's right $\endgroup$ – chickenpie Apr 24 '18 at 13:07
  • $\begingroup$ @Ch Well, the answer then is below. Since you didn't show your way I can't tell where the mistakes are, but your results in the first and last paths are wrong. $\endgroup$ – DonAntonio Apr 24 '18 at 13:08
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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 $\;2\pi i\;$ ...

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  • $\begingroup$ Would it be possible to evaluate the integral using z in the cartesian form, so the integrand would look like (x+iy)(dx+idy), instead of the parameterisation with t? $\endgroup$ – chickenpie Apr 24 '18 at 13:24

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