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Quiz 7 / Re: LEC0101 Quiz#7 ONE-B
« on: December 12, 2020, 04:34:27 PM »
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Quiz 7 / LEC0101 Quiz#7 ONE-B
« on: December 12, 2020, 04:26:52 PM »
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the first quadrant:
$$f(z)=z^9+5z^2+3$$
Answer: First, we separate the line into three parts, denoted (1), (2), and (3).
For (1): $z=x, x: 0 \rightarrow R$ with $R \rightarrow \infty$.
$$f(z) = f(x) = x^9+5x^2+3$$
Thus we have $$arg(f(z))=0$$
Hence, the change of argument is $0$.
For (2): $z=Re^{it}, t: 0 \rightarrow \frac{\pi}{2}$ with $R \rightarrow \infty$.
$$f(z) = R^9e^{i9t}-5R^2e^{i2t}+3 =R^9(e^{i9t}-\frac {5e^{i2t}} {R^7}+\frac{3}{R^9})$$
As $R \rightarrow \infty$, $f(z) \rightarrow R^9e^{i9t}$, where $9t\in [0,\frac {9\pi} {2}]$.
Hence, the change of argument is $\frac {9\pi}{2}$.
For (3): $z=yi, y: R \rightarrow 0$ with $R \rightarrow \infty$.
$$f(z)= iy^9-5y^2+3$$
As $R \rightarrow \infty$, $f(z) \rightarrow iy^{9}$.
As $y = 0$, $f(z) = 3$.
Hence, $f(z)$ traverses from y-axis to $(3,0)$.
Thus, the change of argument is $$0 - \frac{\pi}{2} = \frac{-\pi}{2}$$
Therefore, by definition, $f(z)$ has 2 zeroes in the first quadrant.
$$f(z)=z^9+5z^2+3$$
Answer: First, we separate the line into three parts, denoted (1), (2), and (3).
For (1): $z=x, x: 0 \rightarrow R$ with $R \rightarrow \infty$.
$$f(z) = f(x) = x^9+5x^2+3$$
Thus we have $$arg(f(z))=0$$
Hence, the change of argument is $0$.
For (2): $z=Re^{it}, t: 0 \rightarrow \frac{\pi}{2}$ with $R \rightarrow \infty$.
$$f(z) = R^9e^{i9t}-5R^2e^{i2t}+3 =R^9(e^{i9t}-\frac {5e^{i2t}} {R^7}+\frac{3}{R^9})$$
As $R \rightarrow \infty$, $f(z) \rightarrow R^9e^{i9t}$, where $9t\in [0,\frac {9\pi} {2}]$.
Hence, the change of argument is $\frac {9\pi}{2}$.
For (3): $z=yi, y: R \rightarrow 0$ with $R \rightarrow \infty$.
$$f(z)= iy^9-5y^2+3$$
As $R \rightarrow \infty$, $f(z) \rightarrow iy^{9}$.
As $y = 0$, $f(z) = 3$.
Hence, $f(z)$ traverses from y-axis to $(3,0)$.
Thus, the change of argument is $$0 - \frac{\pi}{2} = \frac{-\pi}{2}$$
Therefore, by definition, $f(z)$ has 2 zeroes in the first quadrant.
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Quiz 5 / LEC0101 Quiz#5 oneC
« on: November 06, 2020, 11:13:23 AM »
Locate each of the isolated singularities of the given function and tell whether it is a removable singularity, a pole, or an essential singularity (case (3)). If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole.
$$\frac{e^z-1}{z}$$
Here is my answer:
$$\frac{e^z-1}{z}$$
Here is my answer:
- First, we get that z=0 is a singularity.
- Next, we have $$\lim_{z\to 0}\frac{e^z-1}{z}=\lim_{z\to 0}\frac{e^z}{1} = \lim_{z\to 0}{e^z} = 1$$
- Hence, by definition, it is a removable singularity.
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Quiz 5 / LEC5101 quiz#5 FiveC
« on: November 05, 2020, 09:20:01 PM »
Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given inhomogeneous equation:
\begin{equation*}
(1-t)y''+ty'-y=2(t-1)^{2}e^{-t}, 0<t<1;
\end{equation*}
\begin{equation*}
y_1(t)=e^{t}, y_2(t)=t.
\end{equation*}
Here is my solution:
\begin{equation*}
(1-t)y''+ty'-y=2(t-1)^{2}e^{-t}, 0<t<1;
\end{equation*}
\begin{equation*}
y_1(t)=e^{t}, y_2(t)=t.
\end{equation*}
Here is my solution:
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Quiz 4 / LEC0101 Quiz#4 oneA
« on: October 23, 2020, 11:27:37 AM »
Evaluate the given integral using the technique of Example 10 of Section 2.3:
$$\int_{γ} \frac{dz}{z^2}$$
where γ is any curve in {z: Re(z)≥0, z≠0}, joining -i to 1+i.
Here is my answer:
First, we observe that γ is not closed.
Since γ is in {z: Re(z)≥0, z≠0}, then
$$f(z)=\frac{1}{z^2}$$ is analytic on D.
Thus we have
$$\int_{γ} \frac{dz}{z^2} = [-\frac{1}{z}]^{1+i}_{-i} = - \frac{1}{1+i}-\frac{1}{i}$$
$$\int_{γ} \frac{dz}{z^2}$$
where γ is any curve in {z: Re(z)≥0, z≠0}, joining -i to 1+i.
Here is my answer:
First, we observe that γ is not closed.
Since γ is in {z: Re(z)≥0, z≠0}, then
$$f(z)=\frac{1}{z^2}$$ is analytic on D.
Thus we have
$$\int_{γ} \frac{dz}{z^2} = [-\frac{1}{z}]^{1+i}_{-i} = - \frac{1}{1+i}-\frac{1}{i}$$
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Quiz 4 / LEC5101 quiz#4 FiveE
« on: October 22, 2020, 07:18:19 PM »
Solve the given equation for t > 0. No need to do an actual change of variables, just use the result of Problem 34 of Homework to Section 3.1:
\begin{equation*}
t^{2}y''+3ty'+1.25y=0.
\end{equation*}
\begin{equation*}
t^{2}y''+3ty'+1.25y=0.
\end{equation*}
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Quiz 3 / Quiz#3 Lec5101 5E
« on: October 08, 2020, 07:18:24 PM »
Hi, this is my solution to the following quiz question.
Find the solution of the given initial value problem.
\begin{equation*}
4y'' - y = 0, y(-2) = 1, y'(-2) = -1.
\end{equation*}
Find the solution of the given initial value problem.
\begin{equation*}
4y'' - y = 0, y(-2) = 1, y'(-2) = -1.
\end{equation*}
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Quiz 2 / Quiz#2 Lec0101 B
« on: October 02, 2020, 11:15:53 AM »
Question: Find the limit of each function at the given point, or explain why it does not exsit.
\begin{equation*}
z_{0}=8+i.
\end{equation*}
Answer:
f(z)=(1−Im z)-1 at z0=8
and then at\begin{equation*}
z_{0}=8+i.
\end{equation*}
Answer:
- When z0 = 8, we have $$\lim_{z\to 8}f(z)=\lim_{z\to 8}\frac{1}{1- Im[8]} = \lim_{z\to 8}\frac{1}{1-0} = 1$$
- When z0 = 8+i, we have $$\lim_{z\to 8+i}f(z)=\lim_{z\to 8+i}\frac{1}{1- Im(8+i)} = \lim_{z\to 8+i}\frac{1}{1-1}$$ Since the denominator cannot be zero, hence when z0 = 8+i, the limit does not exist.
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Quiz 2 / LEC5101 quiz#2 FiveD
« on: October 01, 2020, 07:18:40 PM »
Find the value of b for which the given equation is exact, and then solve it using that value of b.
\begin{equation*}
(ye^{2xy}+x)+bxe^{2xy}y'=0.
\end{equation*}
\begin{equation*}
(ye^{2xy}+x)+bxe^{2xy}y'=0.
\end{equation*}
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Quiz 1 / Quiz#1 LEC0101 1D
« on: September 25, 2020, 11:14:04 AM »
Describe the locus of points z satisfying the given equation:
\begin{equation*}
|z+1|^{2}+2|z|^{2}=|z-1|^{2}
\end{equation*}
Answer:
Let \begin{equation*}
z=x+iy
\end{equation*}
Then, we have
\begin{equation*}
|(x+1)+iy|^{2}+2|x+iy|^{2}=|(x-1)+iy|^{2}
\end{equation*}
Thus,
\begin{equation*}
(x+1)^{2}+y^{2}+2(x^{2}+y^{2})=(x-1)^{2}+y^{2}
\end{equation*}
Simplifying the equation and we get,
\begin{equation*}
x^{2}+2x+y^{2}=0
\end{equation*}
Adding 1 at both sides, we obtain
\begin{equation*}
x^{2}+2x+1+y^{2}=1
\end{equation*}
\begin{equation*}
(x+1)^{2}+y^{2}=1
\end{equation*}
Therefore, the locus of point z is a circle centered at (-1,0) with radius 1.
\begin{equation*}
|z+1|^{2}+2|z|^{2}=|z-1|^{2}
\end{equation*}
Answer:
Let \begin{equation*}
z=x+iy
\end{equation*}
Then, we have
\begin{equation*}
|(x+1)+iy|^{2}+2|x+iy|^{2}=|(x-1)+iy|^{2}
\end{equation*}
Thus,
\begin{equation*}
(x+1)^{2}+y^{2}+2(x^{2}+y^{2})=(x-1)^{2}+y^{2}
\end{equation*}
Simplifying the equation and we get,
\begin{equation*}
x^{2}+2x+y^{2}=0
\end{equation*}
Adding 1 at both sides, we obtain
\begin{equation*}
x^{2}+2x+1+y^{2}=1
\end{equation*}
\begin{equation*}
(x+1)^{2}+y^{2}=1
\end{equation*}
Therefore, the locus of point z is a circle centered at (-1,0) with radius 1.
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Quiz 1 / Quiz#1 Section 5101
« on: September 24, 2020, 08:27:44 PM »
Hi, this is my solution to the following quiz question.
Use the method of variation of parameters to solve the given differential equation:
\begin{equation*}
ty' + 2y = sin(t), t>0.
\end{equation*}
Use the method of variation of parameters to solve the given differential equation:
\begin{equation*}
ty' + 2y = sin(t), t>0.
\end{equation*}
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