Infinite complex nested radical and it's complex conjugate. Today I want to play with $i$ the imaginary unit I have this :
$$\overline{\sqrt{1+i\sqrt{1+i^2\sqrt{1+i^3\sqrt{1+i^4\sqrt{\cdots}}}}}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}\sqrt{1+\frac{1}{i^4}\sqrt{\cdots}}}}}$$
Main remarks we have :
$$\sqrt{1+i\sqrt{1+i^2}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}}}=1$$
$$\overline{\sqrt{1+i\sqrt{1+i^2\sqrt{1+i^3}}}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}}}}$$
$$\overline{\sqrt{1+i\sqrt{1+i^2\sqrt{1+i^3\sqrt{1+i^4}}}}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}\sqrt{1+\frac{1}{i^4}}}}}$$
$$\overline{\sqrt{1+i\sqrt{1+i^2\sqrt{1+i^3\sqrt{1+i^4\sqrt{1+i^5}}}}}}=\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}\sqrt{1+\frac{1}{i^4}\sqrt{1+\frac{1}{i^5}}}}}}$$
And so on...
My question :
What happend in the infinite case ?
Have we a closed form for this special nested radical ? 
Edit : It was in fact the complex conjugate. 
Thanks a lot for sharing your time and knowledge .

Just a comment on this. Well first and foremost, I see that you're doing a lot of radicals of this form & also much trivially we can get; 
$$\sqrt{1+\frac{1}{x}\sqrt{1+\frac{1}{x^2}\sqrt{1+\frac{1}{x^3}\sqrt{1+\frac{1}{x^4}\sqrt{\cdots}}}}}=\frac{1}{x^{2}}\sqrt{x^{4}+\sqrt{x^{6}+\sqrt{x^{8}+\sqrt{x^{10}+\sqrt{x^{12}+...}}}}}$$
So In a way;
$$\sqrt{1+\frac{1}{i}\sqrt{1+\frac{1}{i^2}\sqrt{1+\frac{1}{i^3}\sqrt{1+\frac{1}{i^4}\sqrt{\cdots}}}}}=\frac{1}{i^{2}}\sqrt{i^{4}+\sqrt{i^{6}+\sqrt{i^{8}+\sqrt{i^{10}+\sqrt{i^{12}+...}}}}}$$
$$=-1$$
One can make a sense out of this, and I still don't personally like nested imaginary units myself. But the solution above is most likely wrong, because the identity claimed at first is by my observation only applicable for positive real quantities.
