MAT334--2020F > Test 1

2020TT1 Deferred Sitting #1

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Maria-Clara Eberlein:
In the last line of a) why do we add pi*m (as opposed to 2pi*m)?

How do we find the values in b) ?

A A:
In response to question 1, I think it may be a typo, I have the answer as 2pi*m. 

Unfortunately, I do not have a response for your second question, I am not sure how they found the values.

Firstly, we have the formula $log(z) = ln|z| + i \cdot arg(z)$ and $z:=log(\pm(2n+1)\pi i)$.
Then by plugging $z$ into that formula we have $ln((2n+1)\pi) +  i\cdot arg(\pm(2n+1)\pi i)$.
Note that $\pm(2n+1)\pi i$ is always on the y-axis/imaginary-axis.
Thus $i\cdot arg(\pm(2n+1)\pi i) = i\cdot (\frac{1}{2} + m) \cdot \pi$. Plus we need $m\pi$ here since it is periodic.
Also we will not have $2m\pi$ since it does not include all the case we have in this question. We need to remember we have to include the all y-axis.

About part b of this problem, I did by setting n=0,1,2,... and try m correspondingly. You will see the restriction we have: $D \in {z:|z|<3}$ does not include many cases. Only when $n=0$ and $m=\pm1$ work here (n goes beyond 1 will break our restriction).

Hope it will be helpful :)

Maria-Clara Eberlein:
Thank you! How do we know the period is pi?

Since $\pm (2n+1) \pi i$ can be on both positive y-axis and negative y-axis. Thats why we have m instead of 2m. If we have 2m (period of $2\pi$), we would have either positive or negative y-axis only, which does not include every case.


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