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Discussion Response Post to: Complex no.

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Priyesh (1601)

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i^i^i^i........... = a + ib

=> ${i}^{a+ib}$ = a + ib

=> ${e}^{i(\pi/2)(a+ib)}$ = a + ib

=> $\frac{{e}^{i\pi a/2}}{{e}^{\pi b/2}}$ = a + ib

so mod z = ${e}^{-\pi b/2}$

"Imagination is more important than knowledge."

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