Hint: The expression be equal to \(a+ib\) for some . Then, \(a+ib=i^{(a+ib)}\implies a=e^(-\pi b/2)\cos(a\pi /2), b=e^(−\pi b/2)\sin(a\pi /2)\implies a^2+b^2=e^{-\pi b},\ b=a\tan (a\pi /2)\). One needs to solve the resulting set of transcendental equations.

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TopNewestHint: The expression be equal to \(a+ib\) for some . Then, \(a+ib=i^{(a+ib)}\implies a=e^(-\pi b/2)\cos(a\pi /2), b=e^(−\pi b/2)\sin(a\pi /2)\implies a^2+b^2=e^{-\pi b},\ b=a\tan (a\pi /2)\). One needs to solve the resulting set of transcendental equations.Log in to reply