In Cremona's online book Chapter 2, in order to calculate the lattice invariant we have:
$\tau=\omega_1/\omega_2$
Set $q=e^{2\pi i\tau}$
(2.14.1) $c_4 =(2\pi/\omega_2)^4(1+240 \sum_{n=1}^{\infty} n^3 q^n/(1-q^n))$
and $c_6 =(2\pi/\omega_2)^6(1-504 \sum_{n=1}^{\infty} n^5 q^n/(1-q^n))$
This is supposed to give us integer values for $c_4$ and $c_6$ that we can then use to calculate the discriminant of a modular elliptic curve. But whenever I calculate a few terms of the series with $\tau$ meeting the constraints laid out by Cremona in Chapter 2, it always gives me a complex exponent to $e$. The $i$ never goes away. The $q$ never gives up! How do we end up with the promised integer or rational values for $c_4$ and $c_6$?
Here is an example of what I am seeing:
Let $\omega_1=-3-.8i$
Let $\omega_2=-1+1.6i$
Then, $\tau=\omega_1/\omega_2$ and Cremona's conditions,
$\lvert\tau \rvert=1.64556>1$
$\lvert \Re(\tau) \rvert=.483146<.5$
$\Im(\tau)=1.57303>\sqrt3/2$
are satisfied by our choices. Now instead of an infinite series let's limit it to $5$ terms only. We can calculate $c_4$ only and see the issue.
Set $q=e^{2\pi i\tau}=e^{3.0357i}.000051$. At least that is what the calculator gave for $q$.
$c_4 =(2\pi/\omega_2)^4(1+240 \sum_{n=1}^{5} n^3 q^n/(1-q^n))=e^{-2.23309i}121.48$
If I take 10 terms I get the same result for $c_4$. As you can see, it has an imaginary exponent that I can't get rid of no matter how many terms I use, and that prevents the expected result that Cremona states that $c_4$ will be approach a rational or an integer.
A more general way to ask the question is this:
Since the nome $q$ is complex, then taking one term or a hundred terms in the Lambert series always sums to a complex value, so how can it converge to a rational value? Is the problem the particular $q$ in the example? If so, then how can a $\tau$ that satisfies Cremona's conditions not work?