Complex analysis - integrating over a contour

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November 16th 2012, 06:59 PM
alphabeta89

Complex analysis - integrating over a contour

Let $0 < |\alpha| < 1$ . Find

$\displaystyle{\int_{\gamma}\frac{\mathrm{Re}(z)}{z-\alpha}dz}$

in terms of $\alpha$ , where $\gamma$ is the circle $|z| = 1$ oriented in the counterclockwise
direction.

I tried to express $Re(z)$ as $\frac{z+\overline{z}}{2}$ . Then how should I proceed on? Any idea/suggestion is welcomed!:D
November 16th 2012, 09:37 PM
chiro

Re: Complex analysis - integrating over a contour

Hey alphabeta.

Can you use the Cauchy-Formula and evaluate f(z) at alpha? (Since f(z) has no singularities in the interior of the contour)?
November 16th 2012, 10:12 PM
Prove It

Re: Complex analysis - integrating over a contour

Quote:

Originally Posted by alphabeta89

Let $0 < |\alpha| < 1$ . Find

$\displaystyle{\int_{\gamma}\frac{\mathrm{Re}(z)}{z-\alpha}dz}$

in terms of $\alpha$ , where $\gamma$ is the circle $|z| = 1$ oriented in the counterclockwise
direction.

I tried to express $Re(z)$ as $\frac{z+\overline{z}}{2}$ . Then how should I proceed on? Any idea/suggestion is welcomed!:D

First, we need to check that there are not any singularities. The only place where there may be a singularity is where $\displaystyle \begin{align*} z - \alpha = 0 \implies z = \alpha \end{align*}$ . But this is not possible since on the contour $\displaystyle \begin{align*} |z| = 1 \end{align*}$ , for that denominator to be 0, $\displaystyle \begin{align*} |\alpha| = |z| \end{align*}$ , but we know $\displaystyle \begin{align*} 0 < |\alpha| < 1 \end{align*}$ . Now since $\displaystyle \begin{align*} \gamma \end{align*}$ is the unit circle oriented in the counterclockwise direction centred at the origin, we can parameterise it as $\displaystyle \begin{align*} z = e^{i\,t} \implies dz = i\,e^{i\,t}\,dt \end{align*}$ with $\displaystyle \begin{align*} 0 \leq t \leq 2\pi \end{align*}$ . Notice that $\displaystyle \begin{align*} \textrm{Re}\, (z) = \frac{z + \overline{z}}{2} = \frac{e^{i\,t} + e^{-i\,t}}{2} \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \oint_{\gamma}{\frac{\textrm{Re}\,(z)}{z - \alpha}\,dz} &= \int_0^{2\pi}{\frac{\frac{e^{i\,t} + e^{-i\,t}}{2}}{e^{i\,t} - \alpha} \, i\, e^{i\,t}\, dt} \\ &= \frac{1}{2}\int_0^{2\pi}{\left[\frac{e^{i\,t} + \left( e^{i\,t} \right) ^{-1}}{e^{i\,t} - \alpha} \right] i\,e^{i\,t} \,dt} \end{align*}$

Now let $\displaystyle \begin{align*} u = e^{i\,t} \implies du = i\,e^{i\,t} \, dt \end{align*}$ and note that $\displaystyle \begin{align*} u(0) = 1 \end{align*}$ and $\displaystyle \begin{align*} u(2\pi) = 1 \end{align*}$ giving

$\displaystyle \begin{align*} \frac{1}{2}\int_0^{2\pi}{\left[ \frac{e^{i\,t} + \left( e^{i\,t} \right)^{-1}}{e^{i\,t} - \alpha} \right] i\,e^{i\,t}\,dt } &= \frac{1}{2} \int_1^1{ \frac{ u + u^{-1} }{ u - \alpha } \, du}\\ &= 0 \end{align*}$
November 16th 2012, 10:57 PM
alphabeta89

Re: Complex analysis - integrating over a contour

Quote:

Originally Posted by chiro

Hey alphabeta.

Can you use the Cauchy-Formula and evaluate f(z) at alpha? (Since f(z) has no singularities in the interior of the contour)?

Hi, Cauchy Integral Formula requires f(z) to be analytic rite? But Re(z) is not analytic anywhere right?
November 16th 2012, 11:02 PM
alphabeta89

Re: Complex analysis - integrating over a contour

Quote:

Originally Posted by Prove It

First, we need to check that there are not any singularities. The only place where there may be a singularity is where $\displaystyle \begin{align*} z - \alpha = 0 \implies z = \alpha \end{align*}$ . But this is not possible since on the contour $\displaystyle \begin{align*} |z| = 1 \end{align*}$ , for that denominator to be 0, $\displaystyle \begin{align*} |\alpha| = |z| \end{align*}$ , but we know $\displaystyle \begin{align*} 0 < |\alpha| < 1 \end{align*}$ . Now since $\displaystyle \begin{align*} \gamma \end{align*}$ is the unit circle oriented in the counterclockwise direction centred at the origin, we can parameterise it as $\displaystyle \begin{align*} z = e^{i\,t} \implies dz = i\,e^{i\,t}\,dt \end{align*}$ with $\displaystyle \begin{align*} 0 \leq t \leq 2\pi \end{align*}$ . Notice that $\displaystyle \begin{align*} \textrm{Re}\, (z) = \frac{z + \overline{z}}{2} = \frac{e^{i\,t} + e^{-i\,t}}{2} \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \oint_{\gamma}{\frac{\textrm{Re}\,(z)}{z - \alpha}\,dz} &= \int_0^{2\pi}{\frac{\frac{e^{i\,t} + e^{-i\,t}}{2}}{e^{i\,t} - \alpha} \, i\, e^{i\,t}\, dt} \\ &= \frac{1}{2}\int_0^{2\pi}{\left[\frac{e^{i\,t} + \left( e^{i\,t} \right) ^{-1}}{e^{i\,t} - \alpha} \right] i\,e^{i\,t} \,dt} \end{align*}$

Now let $\displaystyle \begin{align*} u = e^{i\,t} \implies du = i\,e^{i\,t} \, dt \end{align*}$ and note that $\displaystyle \begin{align*} u(0) = 1 \end{align*}$ and $\displaystyle \begin{align*} u(2\pi) = 1 \end{align*}$ giving

$\displaystyle \begin{align*} \frac{1}{2}\int_0^{2\pi}{\left[ \frac{e^{i\,t} + \left( e^{i\,t} \right)^{-1}}{e^{i\,t} - \alpha} \right] i\,e^{i\,t}\,dt } &= \frac{1}{2} \int_1^1{ \frac{ u + u^{-1} }{ u - \alpha } \, du}\\ &= 0 \end{align*}$

I thought we must also check for singularities inside $\gamma$ ?
November 16th 2012, 11:05 PM
Prove It

Re: Complex analysis - integrating over a contour

Quote:

Originally Posted by alphabeta89

Hi, Cauchy Integral Formula requires f(z) to be analytic rite? But Re(z) is not analytic anywhere right?

Last time I checked, all real numbers are analytic...
November 16th 2012, 11:09 PM
Prove It

Re: Complex analysis - integrating over a contour

Quote:

Originally Posted by alphabeta89

I thought we must also check for singularities inside $\gamma$ ?

Actually I think you are correct. Give me a moment...
November 16th 2012, 11:11 PM
alphabeta89

Re: Complex analysis - integrating over a contour

Quote:

Originally Posted by Prove It

Last time I checked, all real numbers are analytic...

No, Re(z) is nowhere analytic rite?
November 16th 2012, 11:17 PM
Prove It

Re: Complex analysis - integrating over a contour

Quote:

Originally Posted by alphabeta89

No, Re(z) is nowhere analytic rite?

You are confusing yourself. z is NOT a function, it is a complex number x + iy, with x and y being real numbers. The real part of z is x. So Re(z) is simply a real number, NOT a function.
November 16th 2012, 11:22 PM
Prove It

Re: Complex analysis - integrating over a contour

To answer your other question, there is a singularity at $\displaystyle \begin{align*} z = \alpha \end{align*}$ inside the region, so you will need to use the residue theorem.

$\displaystyle \begin{align*} \oint {f(z) \, dz} = 2\pi i \textrm{ Res }(f, \alpha) \end{align*}$ , where $\displaystyle \begin{align*} \textrm{Res }(f, \alpha) = \lim_{z \to \alpha} (z - \alpha) \frac{\textrm{Re }(z)}{z - \alpha} \end{align*}$

That limit should be very easy to evaluate...