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**Chapter 1 / Re: Math Notation Question**

« **on:**January 24, 2021, 07:56:05 AM »

it depends: if $u=u(x)$ you can use either. If $u=u(x,y)$ then only partial derivative would be correct.

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it depends: if $u=u(x)$ you can use either. If $u=u(x,y)$ then only partial derivative would be correct.

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You need to indicate that there are no zeroes on $\gamma$

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"How ellipses wouls look like" means the directions and relative size of their semi-axis. See frame 4 of MAT244_W8L3 handout

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There could be misprints

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For higher order equations it is covered in MAT244-LEC0201-W6L2 (see modules). It is mandatory material.

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you need to write it, if you hope for any answer

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Indeed, instead of $\log(\pm w)$ with $+2\pi mi$ we write $\log(w)$ with $+\pi mi$ since $\log (-w)=\log(w)+\i i$

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Of **some** fundamental set (remember a constant factor!)

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Yes, some of them are geometric series, and some of $e^{z}$, $\sin(z)$, $\sinh(z)$ and so on. However some can be derived from those, ether by substitution (f.e. $z^2$ instead of $z$), some by integration, differentiation, multiplication by $z^m$ or combination of both. F.e. consider geometric $\dfrac{1}{1-z}$. Integratinfg we can get power series for $-\Log (1-z)$, diffeerentiating for $\frac{1}{(1-z)^m}$ ,...

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Indeed

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Hi there, I think there might be some problem with Q28. After I found the integrating factor and then solve for the solution, it's impossible to find the solution of my h(y) in this case (see the last line of my writing).If you found an integrating factor correctly there should be no problem to $h(y)$. There is a problem with

In the textbook, it is e^(-2y). Maybe this one is correct.Both are correct

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is du/dx+i dv/dx? Why can't we take the derivative with respect to y?You can take also derivative by $y$ but you need to multiply it by $i$ (think why)

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Try to solve it by yourself (it follows right from the definition)

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why the complex plane is both open and closed? why did it close? Because by definition, a set is called closed if it contains its boundary. And I don't see the complex plane contains its boundary.And what is the boundary of $\mathbb{C}$?

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You should remember that each segment of the polygonal curve may be served not by two discs, but several discs (think why)