I tried to calculate $$\sqrt{ ... {\sqrt{\sqrt i}}}$$ by saying $k =\sqrt{ ... {\sqrt{\sqrt i}}}$ , so the equation $$ \sqrt k = k$$ must be true. When we square each side of the equation, we get$$ k^2 - k = 0$$ so we get $\sqrt{ ... {\sqrt{\sqrt i}}} = 0,1$.
However, I learned that $ai \not = b $ when $a,b \subset \mathbb {R}$, so the equation can't be correct. Where is the flaw in my logic?
On
If you always use the principal square root, the answer is one. $i$ has modulus 1 and argument $\pi/2$. Each time you take the square root, the modulus stays the same and the argument is cut in half. Thus, in the limit you get the complex number with modulus 1 and argument 0, which is otherwise known as 1.
If for $z=r e^{i \theta}$, $\theta \in (-\pi,\pi)$ you define $\sqrt{z}$ by $\sqrt{r} e^{i \frac{\theta}{2}}$ then with: $$u_0=i=e^{i \frac{\pi}{2}}$$ $$u_{k+1}=\sqrt{u_k}$$ then your recurrence is well defined and:
$$u_k=e^{i \frac{\pi}{2^{k+1}}}$$
so indeed $u_k \to e^0=1$.