What is $\sqrt{i^0+\sqrt{i^1+\sqrt{i^2+...}}}$ $$\sqrt{i^0+\sqrt{i^1+\sqrt{i^2+...}}}=?$$
Where $i$ is the imaginary unit, $i=\sqrt{-1}$.
I just want to know whether this is possible or not.
I prefer the limit notation in writing these expressions. Nevertheless, if I have interpreted your notation correctly, then you want something like this:

$L=\sqrt{i^0+\sqrt{i^1+\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+...}}}}}}$

$\implies L^2-1=\sqrt{i^1+\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+...}}}}}$

$\implies (L^2-1)^2-i=\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+...}}}}$

$\implies ((L^2-1)^2-i)^2+1=\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+...}}}$

$\implies (((L^2-1)^2-i)^2+1)^2+i=\sqrt{i^4+\sqrt{i^5+...}}=L$  (Recall that $i^4=1, i^5=i$ and so on)

At least in principle, you can get the answer by solving the equation.

However, remember that complex square roots are multivalued and there is no such thing as a 'positive square root' for complex numbers as used in your notation.

Hence, there still remains some ambiguity in the question and we get multiple answers by this method. https://www.wolframalpha.com/input/?i=solve+%28%28%28L%5E2-1%29%5E2-i%29%5E2%2B1%29%5E2%2Bi%3DL

Hope this helps.
Above equation shown below:

$L=\sqrt{i^0+\sqrt{i^1+\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+...}}}}}}$

I just realized my previous answer does not add up.

Since nested radical's have solution's which are real numbers, 

hence we consider the value of "L" as real.

@tatan gave equation for "L"

$\ (((L^2-1)^2-i)^2+1)^2+i=L$

Solving for $(i)$ we get:

$(i)=[(L^2-1)^8-4(L^2-1)-L]/[(4(L^2-1)^6-1]$  ---(1)

Since right hand side of equation (1) is real &amp; 

the left hand side is imaginary we have a contradiction. 

Hence the reply is negative to "OP" question, if a 

solution is possible to his given equation.