Dear student, The given equation is, iii ∞=x+iyLet iii ∞=nthen we have in=x+iytherefore, in= 1, if n=4q i, if n=4q+1 -1, if n=4q+2 -i, if n=4q+3, q is a natural numberwhen n=4q, then 1=x+iy⇒x=1, y=0⇒x2+y2=1when n=4q+1, then i=x+iy⇒x=0, y=1⇒x2+y2=1when n=4q+2, then -1=x+iy⇒x=-1, y=0⇒x2+y2=1when n=4q+3, then -i=x+iy⇒x=0, y=-1⇒x2+y2=1 Regards

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Dear student,

The given equation is, i^{i^{i^{.^{.^{\infty}}}}} = x + iy Let i^{i^{i^{.^{.^{\infty}}}}} = n then we have i^{n} = x + iy therefore, i^{n} = \{1, if n = 4 q i, if n = 4 q + 1 - 1, if n = 4 q + 2 - i, if n = 4 q + 3, q is a natural number when n = 4 q, then 1 = x + iy \Rightarrow x = 1, y = 0 \Rightarrow x^{2} + y^{2} = 1 when n = 4 q + 1, then i = x + iy \Rightarrow x = 0, y = 1 \Rightarrow x^{2} + y^{2} = 1 when n = 4 q + 2, then - 1 = x + iy \Rightarrow x = - 1, y = 0 \Rightarrow x^{2} + y^{2} = 1 when n = 4 q + 3, then - i = x + iy \Rightarrow x = 0, y = - 1 \Rightarrow x^{2} + y^{2} = 1

Regards

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