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 
 A: $$X=\sqrt{ i\pi+X}$$
$$X^{2}=i\pi+X$$
$$X^{2}-x-i\pi=0$$
$$X_{1}=\dfrac{1+\sqrt{ 1+4i\pi}}{2}$$
$$X_{2}=\dfrac{1-\sqrt{ 1+4i\pi}}{2}$$
$X_{1}$ and $X_{2}$ are not real.
A: Continuing from Jamal Farokhi's answer, since $$X_{1,2}=\dfrac{1\pm\sqrt{ 1+4i\pi}}{2}$$what remains is to explicit the radical.
If $$a+i b=\sqrt{ 1+4i\pi}$$ $$a^2-b^2+2iab=1+4i\pi$$ then $a^2-b^2=1$ and $ab=2\pi$, which by the end give $$X_1=\frac{1}{2}+\frac{1}{2} \sqrt[4]{1+16 \pi ^2} \cos \left(\frac{1}{2} \tan ^{-1}(4
   \pi )\right)+\frac{1}{2} i \sqrt[4]{1+16 \pi ^2} \sin \left(\frac{1}{2} \tan ^{-1}(4
   \pi )\right)$$ $$X_2=\frac{1}{2}-\frac{1}{2} \sqrt[4]{1+16 \pi ^2} \cos \left(\frac{1}{2} \tan ^{-1}(4
   \pi )\right)-\frac{1}{2} i \sqrt[4]{1+16 \pi ^2} \sin \left(\frac{1}{2} \tan ^{-1}(4
   \pi )\right)$$
A: $$
x=\sqrt{i\pi + \sqrt{\cdots}} \\
x^2=i\pi + \sqrt{i\pi + \sqrt{\cdots}} \\
x^2-i\pi = \sqrt{i\pi + \sqrt{\cdots}} = x \\
(x^2-x)^2 + \pi^2 = 0
$$
I am not sure if I can do it this way, it was just how I thought it would work.
A: 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.
A: Let’s call $a=\sqrt{i\pi+\sqrt{i\pi\dots}}$
\begin{align}
a^2 &= i\pi+a \\
a^2-a &= i\pi \\
(a^2-a)^2 &= -\pi^2 \\
(a^2-a)(a^2-a) &= -\pi^2
\end{align}
Solve for $a$.
Hope that helps!
