what is $\sqrt{(i^4)}$ We know that,

$(a^m)^n =  a^{mn} = (a^{n})^m$

So , 


*

*$\sqrt {i^4} = (i^{1/2})^4 = \left(\pm\frac{1+ i}{\sqrt2} \right)^4 = -1$

*$\sqrt {i^4} = (i^{2}) = -1$

*$\sqrt {i^4} = \sqrt1 = 1$
I think only no $3$ is right. I must have violated some rule in $1$ & $2$. Please let me know.
Maybe $(a^m)^n =  a^{mn} = (a^{n})^m$ is true only for Real Numbers.
 A: I would say that $\sqrt{x}$ invokes a choice. It is usually defined as the real number $y\ge0$ such that $y^2 = x$.
So this definition does not carry over to the complex numbers since there is no notion of greater zero.
For general $n\in \Bbb N$ $a^{\frac{1}{n}}$ has $n$ different solutions if you defined it by $\left(a^{\frac{1}{n}}\right)^n= a$, if $n$ is odd and $a$ is real they all agree.
So what happens here could be see as taking different choices.
But as there is no "canonical choice" for $\sqrt{x}$ when $x$ is complex it is usually not defined. What I mean by "canonical choice" for reals it is enough to require $\sqrt{1}=1$ and continuity, but for complex numbers such a function does not exist on the whole complex plane.
For exponents which are not rational this is quite similar: you can define $x^y$ as $e^{y\log x}$ which in the complex case also invokes a choice since $\exp$ is $2\pi$ periodic, i.e. you have to choose a branch of the logarithm. (which is not defined on the whole complex plane)
Back to your question. How does this relate to the rule
$$(a^m)^n = a^{mn} ?$$
In the real case you can make unique choices for both expressions. I will express it with the definition $x^y =e^{y\log x}$. So we have
$$ (a^m)^n = e^{n(log(e^{m\log n}))} = e^{n({m\log n})} =a^{mn} $$
where I just used that $\log \exp x = x$ and for real numbers there is a unique choice for the logarithm. When you want to check that $x^y =e^{y\log x}$ actually makes sense for rational numbers you will see that there is also a choice invoked by defining roots to be positive. But for complex numbers you don't have such a function $\log$.
So for 1 and 2 the rule does not apply and 3 is correct.
A: In $\mathbb C$ we have $\sqrt1=\{1,-1\}$.
Thus, $\sqrt{i^4}=\sqrt1=\{1,-1\}$.
A: $$\sqrt{i^4}=i^2 = -1\tag1$$

Or
$$\sqrt{i^4}= \sqrt{(\sqrt{-1})^4} = (\sqrt{-1})^2 = -1 = i^2 \tag2$$

Or
$$\sqrt{i^4} = \sqrt{i^2\cdot i^2} = \sqrt{i^2}\cdot \sqrt{i^2}= i\cdot i = i^2  = -1\tag3$$
