Cauchy Integral Experiment

[Kouznetsov]

Now you may fit your approximation with function
F_1(z)=L+exp(Lz+R)
at Im(z)>>1, and evaluate the fundamental mathematical constant R...

7.500000000000000*I yields 1.078008533900850 - 0.9465609164056555*I <----

Very good, Mike! My preprint suggests the following value for this R:
1.077961437528 -0.946540963948 I

.. maybe we just have 1.0780 - 0.9465i / 1.0780 - 0.9466i?..

Yes. 5 digits agree.

I think it takes at least 2x as many decimals in the accuracy of the tetration as the target amount of decimals we want for R due to cancellation, round off, etc.

Not really so. I estimate, I got at least 12 digits of R with the complex<double> variables, id est, with the 15 digit arithmetics, but it is some kind of art rather than a science. (I am artist.) With your Simpsons, it will be difficult to do better.
Consider to change for the Gauss-Legendre. How many digits do you need?

Already you may recommend your representation for the complex<float> implementation of the tetrational as confirmed with two independent codes.
At large values of the imaginary part, you may use the asymptotics, it runs faster and provides better precision.

Also, is it time to plot the function on the z plane?

Yes. Go ahead. The float precision should be sufficient for a beautiful zoomable plot.
Does your compiler support any equivalent of the ImplicitPlot function?