Is :$\sqrt{i\pi+\sqrt{i\pi+\sqrt{i\pi+\sqrt\cdots}}}$ irrational or transcendental or real number? Is there someone who can show me if :$$\sqrt{i\pi+\sqrt{i\pi+\sqrt{i\pi+\sqrt\cdots}}}$$ is irrational or real or transcendental number ?
Thank you for any help 

Let's assume this is a complex number $z=x+iy$ in the first quadrant. Then $z+i\pi=z^2$ implies $$x+i(y+\pi)=(x+iy)^2 = x^2-y^2 +2ixy$$ so
$$ x = x^2 - y^2,\ \ \ y + \pi = 2xy.$$
From the second equation we get $x = \frac{y+\pi}{2y}$, which plugging into the first gives
$$ \frac{y+\pi}{2y} = \frac{y^2+\pi^2+2\pi y}{4y^2} - 2y^2,$$
from which 
$$2y(y+\pi) = y^2+\pi^2 + 2\pi y - 4y^4 \Longrightarrow 4y^4 + y^2 -\pi^2=0.$$
Then using the assumption that $y>0$ we obtain
$$ y^2 = \frac{-1 + \sqrt{1+16 \pi^2}}{8} \Longrightarrow y = \frac{\sqrt{-1+\sqrt{1+16\pi^2}}}{2\sqrt{2}}.$$
Computing $x$ using $x= (y+\pi)/2y$ and the above 
$$ x = \frac{1}{2}+\pi\sqrt{\frac{2}{-1+\sqrt{1+16\pi^2}}},$$
so
$$ z = \frac{1}{2}+\pi\sqrt{\frac{2}{-1+\sqrt{1+16\pi^2}}} + \frac{i}{4}\sqrt{-2+2\sqrt{1+16\pi^2}},$$
which is transcendental. The case of the other quadrants should be similar.
