# Complex analysis - integrating over a contour

• Nov 16th 2012, 07:59 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
• Nov 16th 2012, 10:37 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)?
• Nov 16th 2012, 11:12 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*}
• Nov 16th 2012, 11:57 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?
• Nov 17th 2012, 12:02 AM
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 $\gamma$ ?
• Nov 17th 2012, 12:05 AM
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...
• Nov 17th 2012, 12:09 AM
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...
• Nov 17th 2012, 12:11 AM
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?
• Nov 17th 2012, 12:17 AM
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.
• Nov 17th 2012, 12:22 AM
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...