Toronto Math Forum
MAT3342020F => MAT334Tests and Quizzes => Test 1 => Topic started by: MariaClara Eberlein on October 14, 2020, 04:44:15 PM

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) ?

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 $log(z) = lnz + i \cdot arg(z)$ 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 yaxis/imaginaryaxis.
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 yaxis.
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 :)

Thank you! How do we know the period is pi?

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

Oh okay that makes sense thank you so much!

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