Context
Being:
$$K(k)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2{x}}},\tag{1}$$
the complete elliptic integral of the first kind, and
$$K(k')=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k'^2\sin^2{x}}},\tag{2}$$
where $k'=\sqrt{1-k^2}$ is the complementary modulus. It has been considered far in the literature:
$$\frac{K(k'_{n})}{K(k_{n})}=\sqrt{n},\tag{3}$$ where $n$ is an integer. Solutions of the equation $(3)$ are of particular interest. These $k_{n}$ are so called elliptic singular moduli.
If we consider $s=i(\sqrt{2\sqrt{2}+2})$ then it can be proven:
$$K(s)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{2(\sqrt{2}+1)\sin^2{x}+1}}=\frac{\pi^{3/2}\sqrt{\frac{1}{4}-\frac{\sqrt{2}}{8}}}{\Gamma{(5/8)}\Gamma{(7/8)}}\tag{4},$$
and
$$\frac{K(s')}{K(s)}=\frac{\sqrt{2}}{2}+i\tag{5}.$$
Thus considering $q=e^{-\pi K(s')/K(s)}$ and $P(q)=1-24\sum_{n=1}^{\infty}\frac{24n}{1-q^n},$ we have:
$$2P(q^{2})-P(q)=\frac{4K^2(s)}{\pi^2}(1+s^2)$$ which is $$1+24\sum_{n=1}^{\infty}\frac{n(-1)^n}{e^{\pi n/\sqrt{2}}+(-1)^n}=\frac{\pi\left(1- \frac{3\sqrt{2}}{2}\right)}{\Gamma{(5/8)}^2\Gamma{(7/8)}^2}
\tag{6}.$$
As best I know in the literature elliptic singular moduli are real numbers but this case shows that also purely imaginary moduli exists $(k=i(\sqrt{2\sqrt{2}+2}))$ and in these cases $(3)$ has the form of $(5)$ in the sense that $K(k')/K(k)=a+i$ is a complex number with $a$ an algebraic number.
Question
Do you know if someone has considered purely imaginary $k$ for $K(k)$, so that $\frac{K(k')}{K(k)}=a+i$, where $a$ is an algebraic number?
Best Answer
The ideas in your question related to complex values of $K'/K$ have been studied in past. In particular the theory of complex multiplication says that if $\tau$ is an algebraic number of degree $2$ with positive imaginary part and $q=\exp(2\pi i\tau) $ then the value $k=\vartheta_2^2(q)/\vartheta_3^2(q)$ is an algebraic number.
It is not difficult to demonstrate this general theorem from the special case when $\tau=i\sqrt {n} $ with $n$ being a positive integer.
Let us then consider $\tau=i\sqrt{n} $ and $\tau'=(m+i\sqrt{n}) /p$ where $m$ is an integer and $p$ is a positive integer like $n$. Further let $q=\exp(2\pi i\tau), q'=\exp(2\pi i\tau') $. This implies that $q=q'^p$ and hence if elliptic moduli $k, l$ are related to $q', q'^p=q$ respectively then there is a modular equation of degree $p$ connecting $k, l$. And then since $l$ is algebraic, it follows that $k$ is algebraic as well.
However for most typical evaluations of elliptic moduli and integrals it is preferable to make the assumption $k\in(0,1)$ and use formulas like $$K(ik) =\frac{1}{\sqrt{1+k^2}}K\left(\frac{k}{\sqrt {1+k^2}}\right)\tag{1}$$ and $$K(1/k)=k(K(k)\pm iK(k'))\tag{2}$$ whenever needed.
In particular to deal with your elliptic modulus $s=i\sqrt {2\sqrt {2}+2}=ik$ we can observe that $$\frac{k} {\sqrt{1+k^2}}=\sqrt {2(\sqrt{2}-1)}=k'_2$$ and hence $$K(s) =(\sqrt{2}-1)K(k'_2)=\sqrt{2}k_2K(k_2)$$ The value $K(k_2)$ is well known and thus one obtains the desired value of $K(s) $ mentioned in your question.
Next $$s'=\sqrt{1-s^2}=\sqrt{2}+1=1/k_2$$ and hence using $(2)$ we have $$K(s') =k_2(K(k_2)\pm iK(k'_2))=k_2K(k_2)(1\pm i\sqrt {2})$$ Thus we have $$\frac {K(s')} {K(s)} =\frac{1}{\sqrt{2}}\pm i$$ The sign $\pm $ above doesn't matter as $$q_s=\exp(-\pi K(s') /K(s)) =-e^{-\pi/\sqrt{2}}$$ for both the signs. Let $q_2, q'_2$ be the nomes corresponding to moduli $k_2,k'_2$ then we have $q_s=-q'_2$ and $$2P(q_s^2)-P(q_s)=2P({q'} _2^2)-P(-q'_2)=\frac{4K^2(k'_2)}{\pi^2}(1-2{k'} _2^2)$$ The right side equals $$\frac{8K^2(k_2)}{\pi^2}(2k_2^2-1)$$ and is negative thus one can obtain sum $(6)$ of your question as well.
In general I prefer to assume $q\in(0,1)$ as well and if needed I use formulas linking functions of $q$ and $-q$.