if $i$ is a number then what is its numerical value? $ i $ is the unit imaginary part of complex number , but  there is a question which it is mixed me probably i missed the definition of a number , wolfram alpha $ i $ is assumed to be a number , and others assumed it to be  variable because it satisfies $ \sqrt{i^2}$ =$+i$ or  $-i $   then my question here :
Question:
Is $i$  a number then what is it's value ?
 A: Asking what's the value of $i$ is like asking what's the value of $2$. And, just like $i^2$ has two square roots, $i$ and $-i$, $2^2$ has two square roots, $2$, and $-2$.
And yes, it is a number, not a variable.
A: May I give a question to your question : What is a number? You can't expect a rigorous and absolute answer if the question isn't well defined.
Nevertheless, let my give my answer.
i is not a number, in the sense that it doesn't belong to $\mathbb{R}$, the real numbers. 
However, in mathematics we often generalize concrete notions to more abstract ones in order to explain weird phenomena such as that sometimes it is really convenient to assume that $\sqrt{-1}$ does exist and obeys the usual rules of arithmetic in addition to having the property that $(\sqrt{-1})^2=-1$. 
This abstract concept is called a field. More specificaly, $i$ belongs in the field $\mathbb{C}$ of complex numbers.
One can identify $\mathbb{C}$ with the plane $\mathbb{R^2}$  writing :$z=x+iy$ with $x,y \in \mathbb{R}$. In that sense $i=0+1i\cong (0,1)$
Now let  me explain the part that I think causes the confusion:
The plane $\mathbb{R^2}$ consistis of vectors $u=(x,y)$ and we can only add vectors or multiply them by some scalar number. So since we can't multiply vectors with each other it isn't natural to say that vectors are numbers.
On the other hand we can multiply complex numbers and in fact multiplication has all the good properties it has in the Reals, namely it is commutative, associative and distributive.  
So , because they behave like real numbers we also call them numbers. But in reality $\mathbb{C}$ is an extension of $\mathbb{R}$. 
TL;DR:
$\mathbb{C}$ is not an expansion of $\mathbb{R}$ in the sense that $\mathbb{Q}$ is of $\mathbb{Z}$. So it doesn't make sense to ask for the value of $i$ , expecting an answer in the spirit of $π=3.1415...$. Rather it makes (more) sense to ask of its identification with the plane $\mathbb{R^2}$, namely $i=(0,1)$ and its special algebraic relation, namely $i^2=-1$.
A: By definition $i$ is a complex number such that $i^2=-1$.
What about its value?
A similar question could arise for $\sqrt 2$, since $x^2=2$ has not solution over $\mathbb{Q}$, then we extend $\mathbb{Q}$ to $\mathbb{R}$ and define $\sqrt 2$ the real number such that $(\sqrt 2)^2=2$ $\in\mathbb{Q}$.
For $i$ the story is similar, we need to solve $x^2=-1$ which has not solution over $\mathbb{R}$ then we extend $\mathbb{R}$ to $\mathbb{C}$ defining $i$ such that $i^2=-1\in \mathbb{R}$. Since all the algebra works now we can consider the new class of complex numbers $a+ib$.
Thus we can say that we have an imposed (real) value for $i^2$ but not for $i$ that is just a symbol which follows the given rule. Of course, once we have defined the complex numbers we can simply say that the value of $i$ is just $i$ exactly as we do for $\sqrt 2$.
A: Its value is $0+1i$. What else would it be? It's certainly nothing on the number line familiar from a study of $\mathbb{R}$, since no numbers therein square to anything negative; that's why we dream up complex numbers in the first place.
If you're worried dreaming up new beasts is nonsense, that it might lead to a contradiction at any moment, you'll be pleased to know it can be shown if $\mathbb{R}$ makes sense then so is $\mathbb{C}$. There are several ways to do this, each relying on an interpretation of $a+bi$. For example, all the rules of complex numbers follow if we identify this number with the matrix $\left(\begin{array}{cc}
a & b\\
-b & a
\end{array}\right)$. If you do, of course, $i=\left(\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right)$. You could also identify complex numbers with points in $\mathbb{R}^2$, if you add the right operations, so $i=\left(\begin{array}{c}
0\\
1
\end{array}\right)$. Or you could identify complex numbers with polynomials modulo $z^2+1$, so $i=z$.
A: $i$ is a number. But it can't be found on a real number line.
Let's first consider another example:
If I have a number line starting from zero, you can never find $-3$ on it.You can only obtain $-3$ by extending the number line (by flipping it $180^ \circ$). 
Now if you further extend the number line (by flipping it $90^ \circ$), you will get another number line (i.e. imaginary number line) where you can find $i$.

Just like $\sqrt{1^2}=\pm 1$ lies in real number line with $0^ \circ$ rotation and $180^ \circ$ rotation of positive number line, $\sqrt{i^2}=\pm i$ lies in imaginary number line with $90^ \circ$ rotation and $270^ \circ$ rotation of positive number line.
A: It is a complex number, thus a number. 
Depicted as point in the Gauss plane, it is the point with coordinates $(0, 1)$ meaning it has real component $0$ and imaginary component $1$.
Its value is $i$.
Do not let you frighten by the terms "complex" or "imaginary" they are rooted in history and have nothing to do with the every day meaning. 
Those numbers are neither complicated nor non-existing.
