$$J (\theta) = x- \sum_{n=1}^N \sum_{m=1}^M e^{{n}/ \theta }e^{(\frac {i2 \pi n }{M})m }$$
How do I write $J(\theta)$ in terms of magnitude and phase?
P.S $x$ and $\theta$ are real numbers.
Edits: $$J (\theta) = x- \sum_{n=1}^N e^{n/\theta } \sum_{m=1}^M e^{(\frac {i2 \pi n }{M})m }$$ When n = 1; $$ J(\theta) = x - e^{1/\theta } \underbrace{\sum_{m=1}^M e^{(\frac {i2 \pi }{M})m }}_{=0} = x$$ Therefore, $$J (\theta) = x- \sum_{n=2}^N e^{n/\theta } \sum_{m=1}^M e^{(\frac {i2 \pi n }{M})m }$$
Note that $e^{i2\pi n}=1$ Are $x,\theta$ real? If not, there is no clean expression in terms of magnitude and phase.
With the fourth question, you can factor $e^{n/\theta}$ out of the $m$ sum. The sum of the $M^{th}$ roots of unity is $0$, which you can prove by summing the geometric series. Unless $n$ is a multiple of $M$ the double sum is zero. If $n$ is a multiple of $M$, the second exponential is $1$ and the second sum just becomes multiplication by $M$. Now you have a geometric series in $n$ to sum. You still won't know whether the phase of $J(\theta)$ is $0$ or $\pi$ because it depends on the sign of the (real) right hand side.