Author Topic: Term Test 2 sample P1  (Read 1807 times)

Victor Ivrii

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Term Test 2 sample P1
« on: October 30, 2018, 04:56:54 AM »
For $\gamma$ (one of 7 coloured curves on the picture) calculate
$$
\int_\gamma \frac{dz}{z^3-1}.
$$
« Last Edit: October 30, 2018, 05:02:00 AM by Victor Ivrii »

Junya Zhang

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Re: Term Test 2 sample P1
« Reply #1 on: October 30, 2018, 12:40:19 PM »
Deformation Theorem:
Let $\gamma$ be a simple closed curve. Let $D(a, r)$ be an open disc lies inside $\gamma$.
Let $f$ be an analytic function on and inside $\gamma$ except maybe at $a$.
Then $\int_{\gamma} f(z) dz = \int_{|z-a|=r} f(z) dz$

In our question, $f(z) = \frac{1}{z^3 - 1}$, which is analytic everywhere except at the three points labeled on the graph, namely, 1, $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Let A, B, C be three complex numbers such that $f(z) = \frac{A}{z-1} + \frac{B}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} + \frac{C}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}$
Solve for A,B,C we get
$$A = \frac{1}{3}$$ $$B = \frac{2}{-3\sqrt{3}i-3}$$ $$C = \frac{2}{3\sqrt{3}i-3}$$
So, $f(z) = \frac{\frac{1}{3}}{z-1} + \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} + \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}$

Let $\gamma$ be the blue curve.
The Deformation theorem implies that $$\int_{\gamma} \frac{1}{z^3 - 1} dz = \int_{|z-1|=r} \frac{1}{z^3 - 1} dz$$ for some $r>0$.

$$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz
=\int_{|z-1|=r} \frac{\frac{1}{3}}{z-1}
+ \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)}
+ \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)} dz
= \int_{|z-1|=r}\frac{\frac{1}{3}}{z-1} dz
+\int_{|z-1|=r}  \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} dz
+ \int_{|z-1|=r} \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}  dz $$
Since $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$ are outside of the unit circle $|z-1|=r$, then $\int_{|z-1|=r}  \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} dz = 0$ and $\int_{|z-1|=r} \int_{|z-1|=r} \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}  dz =0$ by Cauchy's theorem.
Therefore, $\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \int_{|z-1|=r} \frac{\frac{1}{3}}{z-1} dz = \frac{1}{3}\int_{|z-1|=r} \frac{1}{z-1} dz = \frac{1}{3}\cdot 2\pi i = \frac{2\pi i}{3}$.
Thus, the integral of $f$ on the blue curve is $ \frac{2\pi i}{3}$.

Similarly, for the pink curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz  = \int_{|z-(-\frac{1}{2} + i\frac{\sqrt{3}}{2})|=r} \frac{1}{z^3 - 1} dz =\int_{|z-(-\frac{1}{2} + i\frac{\sqrt{3}}{2})|=r}  \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} dz  = \frac{2}{-3\sqrt{3}i-3} \cdot 2\pi i = \frac{4\pi i}{-3\sqrt{3}i-3}$

For the brown curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz  = \int_{|z-(-\frac{1}{2} - i\frac{\sqrt{3}}{2})|=r} \frac{1}{z^3 - 1} dz = \int_{|z-(-\frac{1}{2} - i\frac{\sqrt{3}}{2})|=r} \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}  dz = \frac{2}{3\sqrt{3}i-3} \cdot 2 \pi i = \frac{4 \pi i}{3\sqrt{3}i-3}$

For the green curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \frac{2\pi i}{3} +  \frac{4\pi i}{-3\sqrt{3}i-3} $

For the orange curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz =  \frac{4\pi i}{-3\sqrt{3}i-3} + \frac{4 \pi i}{3\sqrt{3}i-3}$

For the red curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz =  \frac{2\pi i}{3}  + \frac{4 \pi i}{3\sqrt{3}i-3}$

For the purple curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz =  \frac{2\pi i}{3} + \frac{4\pi i}{-3\sqrt{3}i-3} + \frac{4 \pi i}{3\sqrt{3}i-3}$

This concludes the question.
« Last Edit: October 30, 2018, 01:38:52 PM by Junya Zhang »

Victor Ivrii

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Re: Term Test 2 sample P1
« Reply #2 on: October 30, 2018, 12:44:34 PM »
What about 6 other curves?

Victor Ivrii

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Re: Term Test 2 sample P1
« Reply #3 on: October 30, 2018, 02:10:05 PM »
If you used roots in the form $e^{2\pi i k/3}$, or, if you convert complex numbers to $a+bi$ format, you would definitely got the simpler expressions for all curves, especially going around more than one point.

More precisely, integral over curve cycling all three points will be $0$ and integral over curves, cycling two points equals to minus integral cycling the missed point.
« Last Edit: November 05, 2018, 07:49:07 PM by Victor Ivrii »