I tried finding the value of i, I was stomped.

The whole point of introducing i is to obtain a number that can accomplish things the real numbers couldn't. Which is why you will never find any real number with all the same properties as i, like i²=-1. It's not like π or e, which are special real numbers. It's a completely new thing!
It can be fun to play around with maths! But i has no numerical value in the sense you're looking for. Its value is just i.
i is defined as √(-1). That is its value. It expands out number system into a two-dimensional one.
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1. What do you mean by "find the value"?
2. What do you know about i?

To most mathematicians i is already a number/value and it does not exist on the real line. We know that it *can't* exist on the real line. If you're trying to find the simplest way to express it, then this is (as someone else has said) like trying to 'find the value' of 1, or 3, or 101. It's already in its simplest form. If you're trying to place it somewhere on the real line, then that's not physically possible; it's like trying to write the square root of 2 as a fraction.

Lastly, a note on your logic. You note that it's possible to find a real value for i^(i), and then you assert that if you could find a real x such that x^(x) = i^(i) then you would have x = i. This leap is completely unfounded. It's perfectly possible (trivial and common, even) to perform an operation on two different numbers that gives the same result.

For example:

* I want to find the value of 1.
* I know that 1^(2) \- 3(1) + 2 = 0.
* I also know that 2^(2) \- 3(2) + 2 = 0.
* Is it true that 1 = 2?

If you can see why the above fails, can you see why your logic on i^(i) does too?
i^i means e^(-π/2) right?
you might want to look up something about the complex plane. If you have the normal number line on the x axis, then the y axis will be the imaginary numbers. so i is not on the normal number line its in a whole extra dimension
It might be enlightening/helpful to first define complex numbers. It then becomes easy to define i and see why it behaves the way it does.

You are probably already familiar with real numbers and their operators +, -, \*, and /.  Most students are taught real numbers and their operators fairly early in their education.

Now let's define complex numbers and their operators:

* Define a complex number to be the following: (a,b) where a and b are real numbers.  So, by our definition, the following are all examples of complex numbers: (1,2), (-100,0), (0,pi), (1/2,0.001).  Often a complex number is plotted by plotting the first number a on a horizontal axis and the second number b on a vertical axis.

* Define (a,0) to be equal to the real number a.

* Define the addition of two complex numbers to be the following: (a,b) + (c,d) = (a+c,b+d).

* Define the multiplication of two complex numbers to be the following: (a,b) \* (c,d) = (a\*c-b\*d, a\*d+b\*c).

* Define the symbol i to be equal to the complex number (0,1). So, for example, if b is a real number, we can do the following: i\*b = (0,1)\*(b,0) = (0,b).  The quantity, i\*b, is called an imaginary number.  Also, people often write the complex number (a,b) as a+i\*b.  This is acceptable because a+i\*b = (a,0)+(0,b) = (a,b).

* Finally, define e^i*b to equal cos(b) + i\*sin(b).  This may seem like a strange definition for e^i*b.  If anybody is interested, I can write a few paragraphs explaining *why* this is a good definition.

Using the above definitions, you can now verify that the symbol i agrees with everything you have been taught about imaginary numbers.

For example, i^2 = i\*i = (0,1) \* (0,1) = (-1,0) = -1.

For another example, e^i*pi = cos(pi) + i\*sin(pi) = -1 + i\*0 = -1 + 0 = -1.

As a final example, i^i = ( e^i\*pi/2 )^i = e^i\*i\*pi/2 = e^-pi/2 = ~0.2079.

If you come across anything in complex mathematics that seems mysterious or strange, you can trace it back to these definitions and the mysteriousness will go away.
yes, the solution is x = i. that is the value. it's like trying to find the value of 3. it's just 3.
The value of i is cos(90°)+isin(90°) = e^0.5πi and that also equals √(-1). It is the unit value of the lateral axis or imaginary axis and it is a definition.

The entire math you're doing is basically confirming the value of i as what mathematics defined it as.

And what you mentioned about i being not like any number is true. i is not a real number.
It seems like you’re treating i as a variable like x that has a value. But it’s not a variable, it is a constant. It is a number that satisfies the fact that when you square it, you get -1. I also like to think of it as a unit measure that is situated perpendicular to 1 and -1.

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