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:<math> z = \int_0^{\operatorname{sl} z}\frac{\mathrm{d}t}{\sqrt{1-t^4}} = \int_{\operatorname{cl} z}^1\frac{\mathrm{d}t}{\sqrt{1-t^4}}.</math>
Beyond that square, the functions can be
By comparison, the circular sine and cosine can be defined as the solution to the initial value problem:
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