If $ \sqrt{-i} = \frac {1-i} {\sqrt{2}} \land \sqrt{-i} = i\sqrt{i} \Rightarrow \frac {1-i} {\sqrt{2}} = i\sqrt{i}$
$$\Rightarrow {1-i} = {i}\sqrt{2i}$$
$$\Rightarrow {i}\sqrt{2i} +{i} =1$$
$$\Rightarrow {i}(\sqrt{2i}+1) =1 $$
$$\Rightarrow {1\over{i}} = \sqrt{2i}+1$$
$$\Rightarrow ({i} = {1 \over \sqrt{2i}+1}) $$
How Can That Be Possible?
There's nothing wrong with that except you need to use a non-principal square root of $2i$.
The square roots of $2i$ are $1+i$ and $-1-i$. And indeed, $$ i = \frac{1}{(-1-i)+1} $$
In general the rule $\sqrt a \sqrt b = \sqrt{ab}$ works for complex numbers only if you accept that one or more of the square roots may need to be non-principal.