The value of $i^{i^{i \dots \infty}} = x + iy$ then $x^{2} + y^{2}$ =

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$e^{- πy}$

$- πy$

$πy$

$e^{+ πy}$

answer is A.

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Detailed Solution

$\begin{array}{l} i^{x + iy} = x + iy \Rightarrow e^{i \frac{π}{2} x} \cdot e^{\frac{- π}{2} y} = x + iy \\ \Rightarrow x^{2} + y^{2} = e^{- πy} \end{array}$

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