Is this notation correct? $\sqrt {-25}=±5i$ I know that, It is correct $$\sqrt {-25}=5i$$
I know that, It is wrong $$\sqrt {-25}=-5i$$

Now, Is this notation correct?
$$\sqrt {-25}=±5i?$$

 A: If $z \ne 0$ is a complex number, then there are exactely two distinct numbers $w_1,w_2$ such that 
$w_1^2=z=w_2^2$ (the square-roots of $z$).
If $z=-25$, then $w_1=5i$ and $w_2=-5i$. Sometimes this is written in the form $\sqrt {-25}=±5i$.
A: $\sqrt{-25} = 5i$ is wrong, as is $\sqrt{-25} = -5i$ and $\sqrt{-25} = \pm 5i$. In my opinion, square root symbols do not belong anywhere near numbers apart from non-negative reals, at the very least until you are very familiar with complex arithmetic. They introduce mistakes and misconceptions, and do not really help anything other than marginally shorten some notation. This includes writing $i = \sqrt{-1}$.
What you can say is $(5i)^2 = -25$, or $(-5i)^2 = -25$, or maybe even $(\pm 5i)^2 = -25$. You can also say that the two square roots of $-25$ are $\pm 5i$ (if that is what you automatically think every time you see $\sqrt{-25}$, then you are probably ready to use square root symbols if you really want to).
The defining property of $i$ is written as $i^2 = -1$ instead of the version in the first paragraph. This is how $i$ is formally defined once you get into abstract algebra an field extensions, and it fits perfectly into the algebra students have known for years before they get to complex numbers for the first time, so why it's not the standard approach when we introduce complex numbers is beyond me.
A: $\sqrt{-25}=\sqrt{25}\sqrt{-1}=+5i$
Note: You are generally taking the principal root in this case. In this case, this would be the positive complex root, $5i$.
However, in the equation $$x^2+25=0$$
$$(x+5i)(x-5i)=0$$
The zeros are $\pm5i$.
A: Not everything in mathematics has a universally accepted definition.  To some people, $\sqrt{-25}$ is  defined to be $5i$.  To others, not even $\sqrt{-1}$ is defined.  Instead $i$ is defined to be a choice for one of the two complex numbers such that $i^2=-1$.
