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### Messages - Hyun Woo Lee

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1
##### Term Test 1 / Re: TT1 Problem 4 (morning)
« on: October 19, 2018, 11:35:25 AM »
This is a piece wise curve.
Let’s denote the first curve as $r_1$ And the second curve, line from $3+3i$ to $3$ as $r_2$
Parametrize both curve.
$$r_1(t) = 3e^{it} + 3i, 0 \geq t \geq -\frac{\pi}{2}$$
$$r_2(t) = -3ti + 3 + 3i, 0 \leq t \leq 1$$

Then, $$\int_{L}^{} (z+\bar{z}) dz = \int_{r_1}{} (z+\bar{z})dz + \int_{r_2}{} (z+\bar{z})dz$$

So, $$\int_{r_1} (z+\bar{z})dz =\int_{-\frac{\pi}{2}}^{0} [r(t) + \overline{r(t)}]r’(t) dt = \int_{-\frac{\pi}{2}}^{0} (3e^{it} + 3i + 3e^{-it} -3i)(3ie^{it}) dt$$
Then,
$$\int_{-\frac{\pi}{2}}^{0} \Bigl(9ie^{2it} + 9i \Bigr)dt = \frac{9}{2}e^{2it} + 9it \Big|_\frac{-\pi}{2}^{0}$$
This calculates to $$9 + \frac{9}{2}i\pi$$
Now we have to compute for $r_2$
$$\int_{r_2}{} (z+\bar{z})dz = \int_{0}^{1} [(3+i(3-3t))(3-i(3-3t))](-3i) dt$$
This is $$-3i\int{0}^{1} [1+i(1-t)][1-i(1-t)] dt = -3i\int_{0}^{1} 1 + (1-t)^2 dt = -3i\int_{0}^{1} t^2 - 2t + 2$$
Then,
$$-3i(\frac{t^3}{3} - t^2 + 2t) \Big|_{0}^{1} = -3i(\frac{1}{3} - 1 +2) = -4i$$
Adding the two integrals you get $$9 + \frac{9}{2}i\pi - 4i$$

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##### Thanksgiving bonus / Re: Thanksgiving bonus 1
« on: October 07, 2018, 03:08:47 PM »
Let $$f = u + iv$$
Then, $$u(x, y) = \frac{x}{x^2+y^2}, v(x, y) = \frac{y}{x^2+y^2}$$
Hence, $$\frac{\partial u}{\partial x} = \frac{-(x^2-y^2)}{(x^2+y^2)^2}, \frac{\partial u}{\partial y} = \frac{-2xy}{(x^2+y^2)2}$$
And
$$\frac{\partial v}{\partial x} = \frac{-2xy}{(x^2+y^2)^2}, \frac{\partial v}{\partial y} = \frac{(x^2 - y^2)}{(x^2+y^2)^2}$$
And,
$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0, \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} = 0$$ This means that our function is locally sourceless and irrotational flux on domain D that does not include the origin (where our function is not defined).

Now note that our function can be written as $$f(z) = \frac{1}{\overline{z}}$$
Since $$\frac{1}{\overline{z}} = \frac{x}{x^2+y^2} + i\frac{y}{x^2+y^2}$$
Now, lets take a look on the circle $$|z| = 1$$
The normal component of f is $$f\cdot n = \cos{\theta}\cos{\theta} + \sin{\theta}sin{\theta}$$
Since $$f(z) = \frac {\cos{\theta} +i\sin{\theta}}{r}, r = 1, \theta = arg(\frac{1}{\overline{z}})$$
And $$n = \cos{\theta} - i\sin{\theta}$$
Then,
$$\int_{|z| =1} f\cdot n ds = 1\int_{|z| = 1} ds = 2\pi$$
Hence, our function is locally sourceless and irrotational flow but is not globally sourceless.

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