Given a line from -i to i and a semi circle from -i to i, evaluate the integral over both and explain why they are different

Question

Let gamma 1 be a straight line from -i to i and let gamma 2 be the semi-circle of radius 1 in the right half plane from -i to i.

Evaluate

$$\int_{\gamma_1}f(z)dz$$

and $$\int_{\gamma_2}f(z)dz$$

where f(z)=complex conjugate(z)

And give a reason as to why the two answers are different.

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My approach:

Let $$\gamma_1=(1-t)(-i)+it = i2t-i|t\in[0,1]$$

and

Let $$\gamma_2=e^{i\theta} = \cos(\theta) + i\sin(\theta)|\frac{3\pi}{2}\leq\theta\leq\frac{\pi}{2}$$

Conjugate of $\gamma_1$ $$-i2t+i$$

Conjugate of $\gamma_2$ $$\cos(\theta) - i\sin(\theta)$$

Integrals

$$\int_0^1-i2t+idt=0$$

$$\int_\frac{3\pi}{2}^{\frac{\pi}{2}}\cos(\theta) - i\sin(\theta) = 2$$

Is the above work correct?

Also, would "the integrals are different because the paths/curves were parameterised differently" be a valid reason??

Answer 1

Your approach is on the right path, however you forgot to multiply by the derivative of the curves in the integrals. The general formula for an integral over a parameterized curve is $$ \int_\gamma f(t) \: dt = \int_a^b f(\gamma(t)) \gamma'(t) \: dt $$

Also i would probably chose angles $\theta \in [-\pi/2,\pi/2]$ for the second curve.

Also note that IF $f$ was holomorphic in a simply connected domain containing the two curves, THEN the integrals would have been the same, hence we can conclude that this is not the case (which of course could be verified directly using cauchy-riemann equations).

Answer 2

$f(z) = \bar{z}$ doesn't satisfy the Cauchy Riemann equations so it's not holomorphic. Therefore we can't assume that integrals along different paths give the same value as Cauchy's integral theorem fails.