How can I find the real and imaginary parts of $\sqrt{i+\sqrt{i+\sqrt{i+\dots}}}$? By using $x = \sqrt{i+\sqrt{i+\sqrt{i+\dots}}}$ , I can find a quadratic equation, but how can I separate the real and imaginary parts?
Edit:
My question was how to find the real and imaginary parts of the complex number $\sqrt{i+\sqrt{i+\sqrt{i+\dots}}}$. 
After solving the quadratic equation we'll get the solution - $x_{1,2} = \frac{1}{2}\pm \frac{\sqrt{1+4i}}{2}$. Now the prblem is to find the real and imaginary parts of $\sqrt{1+4i}$. 
Sorry for the confusion.
 A: If we assume that the following transformation makes sense (which makes a few important assumptions to be discussed later):
$$ z = \sqrt{i + \sqrt{i + \sqrt{ i + \ldots}}} = \sqrt{i + z}$$
Let's assume that square roots are OK to use in the context of complex numbers, and ignore some complicating ideas regarding the roots of unity.  We are playing a game with numbers, and having a bit of fun with it.  So we can square both sides then complete the square.
$$\begin{split}
z^2 &= z + i \\
z^2 - z - i &= 0 \\
4z^2 - 4z - 4i &= 0 \\
4z^2 - 4z + 1 - 1 - 4i &= 0 \\
(2z - 1)^2 - (1 + 4i) &= 0 \\
2z - 1 &= \sqrt{1 + 4i} \\
z &= \frac1{2}(1 + \sqrt{1 + 4i}) \\
\end{split}$$
You've probably already made it this far.  (No $\pm$ because we're only considering the positive root for simplicity.)
So let's talk about what it means to take the square root of an imaginary number.
Any complex number can be represented in two forms, a rectangular form and a polar form.  The rectangular form is your standard $a + bi$ that you ought to be familiar with.  Whereas the polar form can be represented as $r \angle \theta$, where $\theta$ is the angle from the positive real axis.
We can convert between the two much in the same manner as standard conversion between rectangular and polar coordinates.  In fact, there's a handy formula that helps with this that's rather famous, it's called Euler's formula.
$$e^{i\theta} = \cos(\theta) + i \sin(\theta)$$
Which the general case of the more famous $e^{\pi i} + 1 = 0$ equation, that you might have heard about.  I'm not going to prove this formula, there are other resources that do this much better than I can using only text.
The bottom line is you can use this formula to represent complex numbers without hiding behind a magic $\angle$ symbol.  Now everything is numbers.  $r\angle \theta = r e^{i \theta}$.
So.  What happens when two numbers are multiplied together?  Well!
$$r_1 e^{i \theta_1} \cdot r_2 e^{i \theta_2} = r_1 r_2 e^{i (\theta_1 + \theta_2)}$$
When two of them are multiplied together, there's two components of the multiplication.  There's a stretching and a rotation.  The stretching occurs in the same manner as real numbers, but the rotation is the part that corresponds to the complex component of the multiplication.  The angles are added together.
Now you may have noticed something interesting.  Because both sine and cosine are periodic functions, more than one representation is possible of the same number.  So bearing that in mind, let's talk about the roots of unity.
What does $\sqrt{i}$ mean?  Well, one interpretation of it is to find a number that when multiplied by itself gives you $i$.  But, we know from real numbers that this isn't necessarily a unique number.  For instance, both -8 and 8 multiply themselves to give 64.  But $\sqrt{64}$ is 8.  There's only one value.  That's because we've decided by convention to take only the positive value of this square root.  But for complex numbers, the only "positive" numbers, are those boring numbers with imaginary part 0 and real part > 0.  So this definition doesn't hold as well.  But let's ignore that for now!  Let's just explore what happens when we try to find numbers.
We want to find $z$ such that $z^2 = i$.  So let's start by expressing $i$ in polar coordinates.
$$i = e^{i \frac{\pi}{2}}$$
This value can be obtained by using Euler's formula.  So we know that whatever number z is, it needs to produce this when multiplied by itself.  Remember that:
$$r_1 e^{i \theta_1} \cdot r_2 e^{i \theta_2} = r_1 r_2 e^{i (\theta_1 + \theta_2)}$$
We can see that $r_1 = r_2 = 1$, and we have $\theta_1 + \theta_2 = \frac{\pi}{2}$, which seems to imply (because $\theta_1 = \theta_2$) that $\theta_1 = \frac{\pi}{4}$.  However, if you remember about that catch where this complex number can have more than one representation.  What if we decide instead to represent $i$ slightly differently.
$$i = e^{i(\frac{\pi}{2} + 2\pi)}$$
Now, this is exactly the same number.  Just a different way of representing it.  However, now when we do the same procedure we get $\theta = \frac{5\pi}{4}$ which is a very different number.  This is why taking the square root is somewhat questionable when dealing with complex numbers.
But hey, we're here to have fun and do some math.  Let's ignore that and say that the "correct" square root is the one performed when $-\pi < \theta \le \pi$.  (This is known as the principal root.) We can do this because we're strong independent people, and to hell with the consequences.
So now let's use this understanding to tackle your question.  How do we find the real part?
$$\begin{split}
z &= \frac1{2}(1 + \sqrt{1 + 4i}) \\
&= \frac1{2}\left(1 + \sqrt{\sqrt{17}e^{i \tan^{-1}(4)}}\right) \\
&= \frac1{2}\left(1 + \sqrt[4]{17}e^{\frac{i}{2} \tan^{-1}(4)}\right) \\
\end{split}$$
Now we can put this back into rectangular form.
$$\begin{split}
z &= \frac1{2}(1 + \sqrt[4]{17}e^{\frac{i}{2} \tan^{-1}(4)}) \\
&= \frac1{2}\left(1 + \sqrt[4]{17}\left(\cos\left(\frac{\tan^{-1}(4)}{2}\right) + i \sin\left(\frac{\tan^{-1}(4)}{2}\right)\right)\right)
\end{split}$$
Which finally gives us our real part that we wanted.
$$\Re(z) = \frac{1 + \sqrt[4]{17}\cos\left(\frac{\tan^{-1}(4)}{2}\right)}{2}$$
(It should be remarked that this number is the same as the one you found below.  $\frac{1 + \sqrt[4]{17}\cos\left(\frac{\tan^{-1}(4)}{2}\right)}{2} = \frac1{2} + \frac{\sqrt{\sqrt{17}+1}}{2\sqrt{2}}$)
Now as a footnotes, let's talk about what assumptions were made to get to the first bit.  What does it actually mean for an infinite nested radical to exist?  Well, one way of looking at it is that it's a value such that the following operation converges on a fixed point. $z_{n+1} = \sqrt{z_n + i}$.  But that begs the question, what is the starting seed value?  (What is $z_0$?)  And from there we can learn a bit more.  For instance, are there any seed values that converge on a different point (for instance, we only considered the principle value of the square root, what occurs when $z_0$ starts at one of the alternate values?  What happens if we chose alternate values of the square root during iteration?  Are there any seed values that fail to converge?
Of course, this is only one way of considering the problem, there are others, which is precisely why other users remarked that the question does not seem well posed.  Which is true, but it is less helpful than it could have been.
A: Consider the recursion
$$x_0=0,\qquad x_{n+1}:=\sqrt{i+x_n}\quad(n\geq0)\ ,$$
where $\sqrt{\cdot}$ denotes the principal value of the square root. It is then easy to see that all $x_n$ $(n\geq1)$ lie in the first quadrant of the complex plane. Solving the equation $x=\sqrt{i+x}$, i.e., $x^2=i+x$,  we obtain a single point in this quadrant, namely
$$\xi={1+\sqrt{1+4i}\over2}=1.30024 + 0.624811 i\ .$$
We now have to prove that the $x_n$ converge to this $\xi$. To this end we consider
$$x_{n+1}-\xi=\sqrt{i+x_n}-\xi={i+x_n-\xi^2\over\sqrt{i+x_n}+\xi}={x_n-\xi\over x_{n+1}+\xi}\ .$$
As both $x_{n+1}$ and $\xi$ lie in the first quadrant we can conclude that $$|x_{n+1}+\xi|>|\xi|>1.3>1\ .$$
This shows that the differences $x_n-\xi$ converge geometrically to $0$, hence $\lim_{n\to\infty}x_n=\xi$.
A: x = $\sqrt{(i+\sqrt{(i+\dots)})}$ = $\sqrt{(i+x)}$
$x^2 - x - i = 0$
Solving for x we get,
$x_{1,2} = \frac{1}{2}\pm \frac{\sqrt{1+4i}}{2}$
Now,
$a + ib =\sqrt{1+4i}$
$\implies a^2 - b^2 + 2iab = 1+4i$
So we have, $a^2-b^2 = 1$ and 2ab = 4.
Solving for a and b we can get
$a = \sqrt(\frac{\sqrt{17}+1}{2})$ and  $b = \sqrt(\frac{\sqrt{17}-1}{2})$.
So the final answer is-
$x = \frac{1}{2} + (\frac{\sqrt{\sqrt{17}+1}}{2\sqrt{2}}) + i (\frac{\sqrt{\sqrt{17}-1}}{2\sqrt{2}})$. 
A: Assuming that we use the principal value of the roots, we find, as you noted, the equation:
$$
\sqrt{i+x}=x \Rightarrow x^2-x-i=0
$$
that has the solution
$$
x=\frac{1\pm\sqrt{1+4i}}{2}
$$
so your problem reduce to find the square root of $1+4i$. Do you know how to do this? 
A: You get $x=\sqrt{x+i}\implies x^2-x-i=0\implies x=\frac{-(-1)\pm\sqrt{1^2-4(1)(-i)}}{2(1)}=\frac{1\pm\sqrt{1+4i}}2$.  Now, $1+4i=\sqrt{17}e^{(\tan^{-1}4 )\cdot i}$.  So $\sqrt{1+4i}=\sqrt[4]{17}e^{\frac{(\tan^{-1}4)\cdot i}2}$.
Finally,  $x=\frac12 \pm\frac{\sqrt[4]{17}e^{\frac{(\tan^{-1}4)\cdot i}2}}2$, from which it follows the real part is $\frac12 \pm\frac{\sqrt[4]{17}}2\cos\frac {\tan^{-1}4}2$.
As to the convergence,  see @Christian Blatter's answer. 
