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### Messages - RunboZhang

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31
##### Quiz 2 / LEC0201 Quiz2 Question2B
« on: October 02, 2020, 01:46:06 PM »
Question and my answer is in the pic attached below.

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##### Chapter 1 / Re: Question on past quiz problem
« on: September 30, 2020, 09:52:48 PM »
Basically you can substitute z by re^i*theta and use the formula Log(z)=ln|z|+iArg(z)=lnr+i*theta where theta belongs to [-pi, pi). After rearranging you will have a sampled equation and it would be much easier to solve. The pic attached below is my solution, correct me if I am wrong.

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##### Chapter 1 / Re: Q10 in Section 1.3
« on: September 30, 2020, 09:30:19 PM »
I think we have restricted arg(z) to (pi/2,pi). After expanding z^2 = (e^i*theta)^2, you will observe that the argument changes to (pi,2pi), which indicates that the second quadrant (excluding axis) is getting mapped to the lower half of the complex plane (still excluding the axises). Thus you can easily prove that this set is open and connected. That is what I think about this question, correct me if I am wrong.

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##### Chapter 1 / Re: how to solve question 19 in section 1.2 in the textbook
« on: September 27, 2020, 09:42:31 AM »
Hello Darren, I have two questions regarding your solution. Firstly, why did you exclude p^2 in equation (1-c^2)x^2-2xp+y^2=0 where p is a real number? And secondly, when you got the solution for eclipse and hyperbola, how did you get 1 on RHS? I thought RHS would be an expression with respect to x and c and it is not necessary to be 1 right?

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##### Chapter 2 / Question on Week3 Lec3 Integrating Factor
« on: September 25, 2020, 04:27:01 PM »
Hi guys, we went through integrating factors in lecture 3 of this week. When using integrating factor to make a non-exact equation exact, we have different types of it. Namely, we have mu(x,y)=mu(x), mu(x,y)=mu(y), and mu(x,y)=mu(xy). I am particularly confused by them. When we are solving a non-exact equation, shall we try three different types and see which one would be helpful, or is there a way to give an instant intuition?

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##### Chapter 1 / Re: past quiz problem
« on: September 24, 2020, 09:38:27 AM »
Hi, the picture attached below is my solution to this problem.

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##### Chapter 1 / Re: how to solve this problem?
« on: September 23, 2020, 06:07:15 PM »
I think there are two ways to solve this problem.

Firstly, you can substitute w by x+iy into the equation. Then organize the equation and put real parts together and imaginary parts together. Now you will have the equation |(x+2)+iy|=|(x-2)+iy|. Then take the square of both sides and organize the equation. Finally you will get x=0, which is the y-axis.

The second way to do this problem is to illustrate it geometrically. |w+2| = |w-2| is same as |w-(-2)|=|w-2|, and it asks you to find all w in complex plane that has equal distances to (2,0) and (-2,0). Thus the answer will be the perpendicular bisector of (2,0) and (-2,0), which is still the y-axis.

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##### Chapter 1 / Question on Textbook Section1.5 The Log Function Example 3
« on: September 22, 2020, 09:16:34 PM »
Hi everyone, I have went through textbook section 1.5 and found an example quite confusing (pic is attached below). In this example, in order to get the limit (n->∞), we discuss the real part and imaginary part separately. But I don't know how does it solve for the real part. In particular, it says we need to use L'Hopital's rule and we will get x as the result. However, I have been stuck at this computation (the draft in orange color is my computation) and it seems like a dead end. I dont know if I did wrong at the very first step. Besides, when we are using L'Hopital's rule here, are we taking the derivative with respect to n or x? How can we know which is the variable?

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##### Chapter 1 / Re: Solving roots of complex numbers
« on: September 20, 2020, 07:18:56 PM »
I think if the angle is unfamiliar, we can leave it as z=r[cos theta + i sin theta], otherwise we need to compute the value of sin and cos. Also, I think it has no difference with z=a+bi, it is just in the polar form.

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##### Chapter 2 / Re: Lec 0101 - 9/15 Question
« on: September 20, 2020, 07:12:43 PM »
Hi, I have plugged it in y and calculated the LHS. Computation is attached below.

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##### Chapter 2 / Re: section 2.1 practice problem 15
« on: September 20, 2020, 10:21:52 AM »
I am sorry, I think there is something I did not make clear. Since at the very first place we have determined x!=(15^(1/2))/2 due to the zero-division error. Beyond that we have initial condition y(2)=0. That indicates that we deduce our ODE from x=(15^(1/2))/2 to x=2. Then you can observe the graph and find its slope is greater than 0 at every defined point. Moreover, since we deduce our ODE starting from (15^(1/2))/2 to 2, we can only go in the same direction due to singularity. I think more will be expanded in lecture.

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##### Chapter 1 / Re: Past quiz 1
« on: September 20, 2020, 09:59:52 AM »
I think it is explained by the first reply in that link. |x+yi-i|=Rez is equivalent to x^2 + (y-1)^2=x^2, subtract both sides by x^2 and this equation would be irrelevant to x. So the only restriction is on y and we have to let y=1. Also, if put it in geometry, it does not matter whether Rez has a positive sign or a negative sign since |x+yi-i| means that the distance between (x,y) and (0,1) is fixed and equal to Rez=x, and it has two corresponding points, (x,1) and (-x,1).

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##### Chapter 2 / Re: section 2.1 practice problem 15
« on: September 19, 2020, 06:51:24 PM »
First of all, it does not equal to 15^(1/2)/2 since the derivative does not exist at x=15^(1/2)/2 (zero devision error). And after you have plotted the curve by part (b) of this question, you would find that y' is greater than 0 at every defined point. Thus only when x>15^(1/2)/2, y>-1/2, its derivative is always positive. Thats why x is strictly greater than 15^(1/2)/2.

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##### Chapter 2 / Re: LEC 0201 Question
« on: September 19, 2020, 02:59:05 PM »