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### Messages - Heng Kan

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##### End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 3
« on: December 02, 2018, 05:24:08 PM »
So is there anything wrong with my calculation? Thanks.

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##### End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 3
« on: December 02, 2018, 04:10:22 PM »
For the residues, please see the attatched scanned pictures.  Also, as for infinity, since infinity is not an isolated singularity, the residue at infinity is not defined.

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##### Quiz-7 / Re: Q7 TUT 0202
« on: December 01, 2018, 06:01:19 PM »
Please see the new attached scanned picture.   For yi on Ri to 0, as long as y is positive, f(yi) always lies in the fourth quardarnt.When R tends to be infinity, Re(f(iR)) = 7 and Im(f(iR)) tends to be negtive infinity. When R=0, f(iR) = 7. So f(iy) approximately rotates from negative imaginary axis to positive real axis counter-clockwisely.

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##### Quiz-7 / Re: Q7 TUT 0202
« on: December 01, 2018, 03:40:52 PM »
Please see the new attached scanned picture.   For yi on Ri to 0, as long as y is positive, f(yi) always lies in the fourth quardarnt.When R tends to be infinity, Re(f(iR)) = 7 and Im(f(iR)) tends to be negtive infinity. When R=0, f(iR) = 7. So f(iy) approximately rotates from negative imaginary axis to positive real axis counter-clockwisely.

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##### Quiz-7 / Re: Q7 TUT 5301
« on: November 30, 2018, 05:00:10 PM »
Si Ying, I think there is somthing wrong in your solution, although your answer is correct.  You cannot say |z^3+1|=|1^3+1| when
|z| = 1 and |z^3+1| = |2^3+1| when |z|=2 because z^3 is not the same as 1^3 or 2^3. They just have the same modulus.
See my answer on the scanned picture.

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##### Term Test 2 / Re: TT2A Problem 4
« on: November 25, 2018, 12:17:50 PM »
There are four scanned pictures.  The first one is for part a, the second and third one are for part b, the last one is for part c.

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##### Quiz-6 / Re: Q6 TUT 0202
« on: November 17, 2018, 05:30:38 PM »
I think you got the coefficients correctly but the Laurent series  is wrong.  Here is my answer. See the attatched scanned picture.

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##### Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P4
« on: November 12, 2018, 01:30:48 PM »
I think it is OK  this time. There are 4 pictures. The first one is for part(a), the second and third one are for part(b), and the last one is for part (c)

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##### Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P4
« on: November 11, 2018, 09:28:46 PM »
As this question is a bit troublesome, there are two scanned pictures.  The final answer is π^3/8, which is shown in the last line of the second picture.

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##### Quiz-5 / Re: Q5 TUT 5201
« on: November 02, 2018, 07:46:50 PM »
See the attatched scanned picture. A different way to get the expansion.

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##### Quiz-4 / Re: Q4 TUT 0102
« on: October 26, 2018, 06:23:21 PM »
See the attatched scanned picture. And the question should add an term which is 1/2pi

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##### Term Test 1 / Re: TT1 Problem 2 (night)
« on: October 19, 2018, 09:45:26 AM »
I think the question means that if the radius of convergence is positive,you have to figure out whether the series is convergent at the radius of convergence. It doesn't mean the radius is always positive.

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##### Term Test 1 / Re: TT1 Problem 2 (night)
« on: October 19, 2018, 09:34:01 AM »
See the attached scanned picture.

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##### Quiz-3 / Re: Q3 TUT 0202
« on: October 12, 2018, 06:24:31 PM »
See the attached scanned picture.

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##### MAT334--Lectures & Home Assignments / Section 1.4 Question 12
« on: October 02, 2018, 11:11:36 PM »
For this question, we have to compute the limit of $f(z) = (z-2)\log|z-2|$ at z=2. As z approaches to $2$, $z-2$ approaches to $0$ and $\log|z-2|$ approaches to negative infinity. If this is a real function, we can use the L'Hopital's rule to compute the limit. So in this situation where it is a complex function, how do we find the limit?  Thanks.

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