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### Messages - Yuyan Liu

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##### Quiz 4 / Quiz4 Lec5101 5E
« on: November 18, 2020, 09:40:37 PM »
Problem: Evaluate the given integral using the technique of Example 10 of Section 2.3:

$$\int_\gamma \frac{dz}{z^3} ,$$

where $\gamma$ is any curve in $\{z: Re z \geq 0, z \neq 0 \}$, joining $-i$ to $1 + i$.

since f is analytic in all $Re(z) \geq 0, z \neq 0, \int_\gamma \frac{dz}{z^3} = F(1+i) - F(-i)$

$$F'(z) = f(z) \Rightarrow F(z) = - \frac{1}{2z^2}$$
$$F(1+i) = - \frac{1}{2(1+i)^2} = - \frac{1}{4i}$$
$$F(-i) = - \frac{1}{2(-i)^2} = \frac{1}{2}$$

Thus, $\int_\gamma \frac{dz}{z^3} = F(1+i) - F(-i) = - \frac{1}{4i} - \frac{1}{2} = \frac{2i-1}{4i}$

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##### Test 2 / Question 3c for 2020S Night Sitting
« on: October 27, 2020, 06:19:33 PM »
Can someone explain how to get to the answer of $cosh(z)$?

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##### Quiz 2 / Quiz2 Lec5101 5D
« on: October 03, 2020, 05:03:08 PM »
Question: Find the limit at $\infty$ of the given function, or explain why it does not exist:
$g(z) = \frac {4 z^6 - 7 z^3}{(z^2 - 4)^3}$

$g(z) = \frac {4 z^6 - 7 z^3}{z^6 + 3(-4) z^4 + 3(-4)^2 z^2 + (-4)^3} = \frac {4 z^6 - 7 z^3}{z^6 - 12 z^4 + 48 z^2 - 64}$
Divide both denominator and numerator by $z^6$:
$g(z) = \frac {4 - 7 z^{-3}}{1 - 12 z^{-2} + 48 z^{-4} - 64 z^{-6}}$
$$\lim_{z \to\infty} g(z) = \frac {4 - 7 \infty ^{3}}{1 - 12 \infty ^{-2} + 48 \infty ^{-4} - 64 \infty ^{-6}} = \frac {4 - 0}{1 - 0 +0 - 0} = 4$$