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To justify: let us change continuously matrix $A$, so that eigenvalues remain complex conjugate, then it may change from stable to unstable but still remain focal point/center with the same direction of rotation. Then we can make $a=d=0$ and $b=-c=\pm 1$ resulting in the system
$$
x'=b y,\\
y'=-bx
$$which solves to $$x=R\cos (bt+ \phi_0),\\
y=R\sin (bt+\phi_0)
$$
or in the polar coordinates
$$r =\mathrm{const},\ \theta = bt +\phi_0$$
which describes a circle with the counter-clockwise (as $b<0$) or clockwise  (as $b>0$) rotation.

It is easy to remember: $x=\cos(t),\ y=\sin(t)$ is counter-clockwise  and satisfies $x'=-y,\ y'=x$.
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Quiz-6 / Re: Q6 TUT 0401
« Last post by Michael Poon on Today at 03:29:21 AM »
Thank you for posting the announcement about the explanation of CW/CCW. I found the textbook a little confusing to read and myself and a few others found this video helpful: https://www.youtube.com/watch?v=dpbRUQ-5YWc

At time 19:42 they display a technique to determine CW vs CCW using generic vectors and matrix A. I think it might be more intuitive but not as rigorous as the explanation you gave. And you explanation you gave in the announcement was very helpful!
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Quiz-6 / Re: Q6 TUT 0401
« Last post by Victor Ivrii on Today at 02:15:43 AM »
Michael, I asked why it is clockwise. Not about stability.

I especially made an announcement. For not providing explanation about clockwise/counter-clockwise rotation on Test (and Exam) the mark will be reduced
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MAT334--Lectures & Home Assignments / Re: 2.5 Q19
« Last post by Victor Ivrii on Today at 02:12:25 AM »
Ende Jin
Please provide exact citation from the textbook.
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MAT334--Lectures & Home Assignments / Re: 2.5 Q19
« Last post by Ende Jin on November 18, 2018, 11:58:09 PM »
I realized that you provided a solution different from the given one back in the book.
But the question doesn't say $\sum a_n(z-z_0)^n = 0$ when $r < |z-z_0| < R$. It just converges. The series converges to zero only if $0 < |z - z_0| < r$.
That is basically my question. The first condition doesn't seem helpful.
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Quiz-6 / Re: Q6 TUT 0401
« Last post by Michael Poon on November 18, 2018, 11:30:36 PM »
Our graphs are all clockwise :)
Why?

If we choose a unit vector $\begin{bmatrix}1 \\ 0\end{bmatrix}$ and do matrix vector multiplication with the matrix $\begin{bmatrix} \alpha > 0 &  1 \\ -1 & \alpha > 0 \end{bmatrix}$, the vector $\begin{bmatrix}\alpha > 0 \\ -1\end{bmatrix}$ follows the phaseportrait CW. $\alpha$ > 0 means the phaseportrait points outward and is unstable.

If we choose a unit vector $\begin{bmatrix}1 \\ 0\end{bmatrix}$ and do matrix vector multiplication with the matrix $\begin{bmatrix} \alpha < 0 &  1 \\ -1 & \alpha < 0 \end{bmatrix}$, the vector $\begin{bmatrix}\alpha < 0 \\ -1\end{bmatrix}$ also follows the phaseportrait CW. $\alpha$ < 0 means the phaseportrait points inward and is stable.
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Quiz-6 / Re: Q6 TUT 0401
« Last post by Michael Poon on November 18, 2018, 11:18:17 PM »
Our graphs are all clockwise :)

Isn't Guanyao's 2nd graph counterclockwise?
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Quiz-6 / Re: Q6 TUT 0401
« Last post by Nikita Dua on November 18, 2018, 11:10:01 PM »
The potraits are clockwise, Since b > 0 and  c < 0
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Quiz-6 / Re: Q6 TUT 0401
« Last post by Victor Ivrii on November 18, 2018, 10:59:34 PM »
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Quiz-5 / Re: Q5 TUT 0501
« Last post by Guanyao Liang on November 18, 2018, 08:30:26 PM »
I think Michael's x2 is wrong. I also think c1 =-2 and c2 =2 same answer with Zihan.
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