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Messages - ZhenDi Pan

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1
Term Test 2 / Re: TT2A Problem 1
« on: November 24, 2018, 06:23:00 AM »
Yes thank you I corrected it. Still our answers are different, I don't know where went wrong though.

2
Term Test 2 / Re: TT2A Problem 1
« on: November 24, 2018, 05:30:25 AM »
We have

\int_\Gamma \frac{zdz}{z^2-4z+5}

Let

f(z) = \frac{z}{z^2-4z+5} = \frac{z}{(z-(2-i))(z-(2+i))}

Question a:
The point $2-i$ is outside of the contour $\Gamma$ and the point $2+i$ is inside of the contour $\Gamma$. Then let

g(z) =\frac{z}{z-2+i} \\
g(2+i) = \frac{2+i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(2+i))}dz = 2\pi i g(z_0) = 2\pi i g(2+i) = 2\pi i \cdot \frac{2+i}{2i}= \pi(2+i)

Question b:
The point $2+i$ is outside of the contour $\Gamma$ and the point $2-i$ is inside of the contour $\Gamma$. Then let

g(z) =\frac{z}{z-2-i} \\
g(2-i) = -\frac{2-i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(2-i))}dz = 2\pi i g(z_0) = 2\pi i g(2-i) = 2\pi i \cdot -\frac{2-i}{2i}= -\pi(2-i)

Question c:
Both points $2+i$ and $2-i$ are inside of the coutour $\Gamma$. Then we have

z_0 = 2+i \\
z_1 = 2-i \\
\left.Res(f;2+i) = \frac{z}{z-2+i} \right\vert_{z=2+i} = \frac{2+i}{2i} \\
\left.Res(f;2-i) = \frac{z}{z-2-i} \right\vert_{z=2-i} = - \frac{2-i}{2i}

So the Residue Theorem gives us

\int_\Gamma f(z)dz = 2\pi i(\frac{2+i}{2i}-\frac{2-i}{2i}) = 2\pi i \cdot 1= 2\pi i

3
Term Test 2 / Re: TT2 Problem 1
« on: November 24, 2018, 04:49:12 AM »
We have

\int_\Gamma \frac{dz}{z^2-2z+10}

Let

f(z) = \frac{1}{z^2-2z+10} = \frac{1}{(z-(1-3i))(z-(1+3i))}

Question a:
The point $1-3i$ is outside of the contour $\Gamma$ and the point $1+3i$ is inside of the contour $\Gamma$. Then let

g(z) =\frac{1}{z-1+3i} \\
g(1+3i) = \frac{1}{6i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(1+3i))}dz = 2\pi i g(z_0) = 2\pi i g(1+3i) = 2\pi i \cdot \frac{1}{6i}=\frac{\pi}{3}

Question b:
The point $1+3i$ is outside of the contour $\Gamma$ and the point $1-3i$ is inside of the contour $\Gamma$. Then let

g(z) =\frac{1}{z-1-3i} \\
g(1-3i) = -\frac{1}{6i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(1-3i))}dz = 2\pi i g(z_0) = 2\pi i g(1-3i) = 2\pi i \cdot -\frac{1}{6i}= -\frac{\pi}{3}

Question c:
Both points $1+3i$ and $1-3i$ are inside of the coutour $\Gamma$. Then we have

z_0 = 1+3i \\
z_1 = 1-3i \\
\left.Res(f;1+3i) = \frac{1}{z-1+3i} \right\vert_{z=1+3i} = \frac{1}{6i} \\
\left.Res(f;1+3i) = \frac{1}{z-1-3i} \right\vert_{z=1-3i} = - \frac{1}{6i}

So the Residue Theorem gives us

\int_\Gamma f(z)dz = 2\pi i(-\frac{1}{6i}+\frac{1}{6i}) = 2\pi i \cdot 0 = 0

4
Term Test 2 / Re: TT2 Problem 2
« on: November 24, 2018, 04:47:23 AM »
For question a, we have

f(z)=(1-z)^{-1/2} \\
a_n = \frac{f^{(n)}(z_0)}{n!} =  \frac{f^{(n)}(0)}{n!}

Then the $nth$ derivative of $f(z)$ can be derived as

f^\prime(z) = \frac{1}{2}(1-z)^{-3/2} \\
f''(z) = \frac{3}{4}(1-z)^{-5/2} \\
f'''(z) = \frac{15}{8} \times  (1-z)^{-7/2} \\
f''''(z) =\frac{105}{16} \times (1-z)^{-9/2}

At $z=0$

f(0) = 1
f'(0) = \frac{1}{2} \\
f''(0) = \frac{3}{4} \\
f'''(0) = \frac{15}{8} \\
f''''(0) =  \frac{105}{16} \\
f^{(n)}(0) =  \frac{1 \times 3 \times \dots \times (2n-1)}{2^n} \\
a_n = \frac{1 \times 3 \times 5 \times \dots \times (2n-1)}{2^n n!}

Thus we have the power series

f(z)= \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n = 1 + \frac{z}{2} + \frac{3z^2}{8}+ \dots

\frac{1}{R} = \lim_{n \to \infty} |\frac{a_{n+1}}{a_{n}}| \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{f^{(n+1)}(0)}{(n+1)!} \times \frac{n!}{f^{(n)}(0)} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{2n+1}{2(n+1)} \times \frac{1}{1} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{2n+1}{2n+2} \mid = 1 \\
R = 1

For question b, let

F(z) = \arcsin(z) \\
F'(z)=\frac{1}{\sqrt{1-z^2}} = (1-z^2)^{-1/2}

Note that

f(z^2)=(1-z^2)^{-1/2} \Rightarrow F'(z)=f(z^2) \\
F(z) = \int f(z^2)

Then

f(z^2) = \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n-1)}{2^n n!}z^2n \\
F(z) = \int f(z^2) = \int \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n+1)}{2^n n!}z^{2n} dz
F(z) = (\sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n-1)}{2^n n! \cdot (2n+1)}z^{2n+1}) +C

Since $F(0) = 0 \Rightarrow C=0$

F(z) = \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n+1)}{2^n n! \cdot (2n+!)}z^{2n+1}

5
Quiz-5 / Re: Q5 TUT 0202
« on: November 02, 2018, 05:06:37 PM »
We have

\int_{0}^{1/2} \log(1 - t z) dt

Integrate over $\gamma$ with respect to $z$, consider

f(z)=\log(1-tz)

Function $f(z)$ is analytic on $\mid z \mid <2$, by Cauchy's theorem, for any closed curve $\gamma$

\int_\gamma f(z)dz = 0 \\
\int_{0}^{1/2} (\int_\gamma \log(1-zt) \,dz)\,dt = \int_{0}^{1/2} 0\,dt

So it is analytic on domain $\mid z\mid < 2$.

Since $\log(1 - t z) = \sum_{n=1}^\infty \frac{-( z t)^n}{n}$ is valid when $\mid zt \mid<1$, and since $\mid z \mid<2$, for all  $t \in [0,\frac{1}{2}]$. We have

\int_{0}^{1/2} \log(1 - t z) dt  = - \int_{0}^{1/2} \sum_{n=1}^\infty  \frac{( z t)^n}{n} dt \\
= - \sum_{n=1}^\infty  \int_{0}^{1/2}  \frac{( z t)^n}{n} dt \\
=  -\sum_{n=1}^\infty\frac{1}{2^{n+1} n (n+1)} z^n

6
Quiz-5 / Re: Q5 TUT 5201
« on: November 02, 2018, 04:52:51 PM »
We can first rewrite $\frac{z+2}{z+3}$ as

\frac{z+2}{z+3} = \frac{z+3-1}{z+3}

\begin{align*}
\frac{z+3-1}{z+3} & = 1 - \frac{1}{z+3} \\
& = 1 - \frac{1}{z+1+2} \\
& = 1 - \frac{1}{2} \times \frac{1}{1+\frac{(z+1)}{2}} \\
& = 1 - \frac{1}{2} \times \frac{1}{1-\frac{-(z+1)}{2}}
\end{align*}
The power series expansion is

1 - \frac{1}{2} \times \frac{1}{1-\frac{-(z+1)}{2}} = 1 - \frac{1}{2} \times \sum_{n=0}^{\infty} (\frac{-(z+1)}{2})^n \\
= 1 - \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(z-(-1))^n = 1 - \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(z+1)^n

The largest disc in which the series is valid:

\mid -\frac{z+1}{2}\mid < 1 \\
\mid z+1\mid < 2

7
Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P2
« on: November 01, 2018, 10:11:40 PM »
First we have

f(z)=2(1-z)^{1/2}

Then the $nth$ derivative of $f(z)$ can be derived as

f^\prime(z) = -(1-z)^{-1/2} \\
f''(z) = -\frac{1}{2}(1-z)^{-3/2} \\
f'''(z) = -\frac{1}{2} \times  \frac{3}{2}(1-z)^{-5/2} \\
f^{(n)}(z) =-(\frac{1}{2} \times \frac{3}{2} \times \dots \times \frac{2n-3}{2})(1-z)^{-(2n-1)/2}

At $z=0$

f(0) = 2
f'(0) = -1 \\
f''(0) = -\frac{1}{2} \\
f'''(0) = -\frac{1}{2} \times \frac{3}{2} \\
f''''(0) =  -\frac{1}{2} \times \frac{3}{2} \times \frac{5}{2} \\
f^{(n)}(0) =   -\frac{1}{2} \times \frac{3}{2} \times \dots \times \frac{2n-3}{2}

Thus we have the power series

f(z)= \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n = 2 - z -\frac{1}{4}z^2 - \dots

\frac{1}{R} = \lim_{n \to \infty} |\frac{a_{n+1}}{a_{n}}| \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{f^{(n+1)}(0)}{(n+1)!} \times \frac{n!}{f^{(n)}(0)} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{2n-1}{2(n+1)} \times \frac{1}{1} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{2n-1}{2n+2} \mid = 1 \\
R = 1

8
Quiz-4 / Re: Q4 TUT 5102
« on: October 26, 2018, 05:59:31 PM »
First we have

\int\limits_{\mid z \mid = 2}\frac{e^z}{z(z-3)}\,dz

The point $3$ is outside of the circle $\mid z \mid = 2$ and the point 0 is inside of the circle $\mid z \mid = 2$. Hence

\int\limits_{\mid z \mid = 2}\frac{e^z}{z(z-3)}\,dz  = \int\limits_{\mid z \mid = 2}\frac{\frac{e^z}{z-3}}{z}\,dz

This gives us function $f(z)$

f(z) = \frac{e^z}{z-3} \Rightarrow f(0) = -\frac{1}{3}

So Cauchy's Formula gives us

\int\limits_{\mid z \mid = 2}\frac{e^z}{z(z-3)}\,dz =2\pi i\frac{e^0}{0-3}= -\frac{2\pi i}{3}

See the attachment for the circle in this question.

9
Term Test 1 / Re: TT1 Problem 1 (night)
« on: October 19, 2018, 05:55:03 AM »
My solution:
\begin{align*}
| \sin(z)^{2}| & =[\sin (x) \cos(iy)]^{2}+[\cos (x)\sin(iy)]^{2}\\
& = [\sin^{2}(x)\cosh^{2}(y)]+[\cos^{2}(x)\sinh^{2}(y)] \\
& = [\sin^2(x)(1+\sinh^2(y))]+[(1-\sin^2(x))\sinh^2(y)] \\
& = \sin^2(x)+\sinh^2(y)-\sin^2(x)\sinh^2(y)+\sin^2(x)\sinh^2(y) \\
& = \sin^2(x)+\sinh^2(y)
\end{align*}

10
Term Test 1 / Re: TT1 Problem 1 (noon)
« on: October 19, 2018, 05:31:45 AM »
My solution:
\begin{align*}
&\cos(z)=\frac{1}{2}(e^{iz}+e^{-iz})=3 \\
&(e^{iz}+e^{-iz}) = 6 \\
&e^{2iz}+1 =6e^{iz} \\
&e^{2iz}+1-6e^{iz} =0 \\
&e^{iz} = \frac{6\pm \sqrt{36-4}}{2}=3\pm 2\sqrt{2} \\
&iz=\log(3\pm 2\sqrt{2}) \\
&iz=\log(3\pm 2\sqrt{2})=\ln(3\pm 2\sqrt{2})+i2k\pi \qquad  k\in \mathbb{Z} \\
&z = -i\ln(3\pm 2\sqrt{2})+2k\pi \qquad k \in \mathbb{Z}
\end{align*}

Please reformat: \cos, \sin , \ln, \log etc

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