\tan^{-1}{\left(\frac{2z}{z^2-1}\right)}=2\tan^{-1}{\left(\frac{1}{z}\right)} \pm n\pi
Here n is an integer. Just saying ... if you solve the problem the way I demonstrate in class (which is, I think, conceptually easier and leads to the appropriate limits more easily), there is some work involved to check that is the same as Griffiths' result.
I'm sure you realized that you can't use Gauss' law by this point. The fields of a finite square plate have an icky symmetry to them, as does anything square-ish when you're dealing with radial fields.
Also, problem 10 is the nearly same as a PH106 problem I assigned last year. Excepting that the integrations involved are more painful.
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