Let $z=r(\cos \theta+i\sin \theta)$
$z^{-1}=r^{-1}(\cos \theta-i\sin \theta)$
We have ,
$|z+\frac{1}{z}|=|(r+1/r)\cos \theta +(r-1/r)i\sin\theta|$
So we have,
$\displaystyle|(r+1/r)\cos \theta +(r-1/r)i\sin\theta|=\sqrt{(r^2+1/r^2)+2\cos 2\theta}$
Thus we have ,
$(r^2+1/r^2)+2\cos 2\theta=a^2$
Let $r^2=x$,
$\Rightarrow x^2+1=x(a^2-2\cos2\theta)$
$\Rightarrow x^2-x(a^2-2\cos2\theta)+1=0$
Solving the quadratic we have,
$\displaystyle x=\frac{(a^2-2\cos 2\theta)+\sqrt{(a^2-2\cos 2\theta)^2-4}}{2}$ and $\displaystyle x=\frac{(a^2-2\cos 2\theta)-\sqrt{(a^2-2\cos 2\theta)^2-4}}{2}$
Clearly x attains its maximum value at $2\theta=0$ using the first expression,
The value is $\displaystyle \frac{(a^2+2)+\sqrt{(a^2+2)^2-4}}{2}$
So the maximum value of $r=\sqrt{\displaystyle \frac{(a^2+2)+\sqrt{(a^2+2)^2-4}}{2}}$
This on simplification turns out to be, $\displaystyle r=\frac{a+\sqrt{a^2+4}}{2}$
Minimum value can also be obtained in that way.
The definition (or one possible definition) of the absolute value of the complex number $a+bi$ (where $a$ and $b$ are real) is $\sqrt{a^2+b^2}$. So you ask, why is it not $\sqrt{a^2+(bi)^2}$ instead? The answer is that this is simply how we choose to define it. You could define a different quantity which is $\sqrt{a^2+(bi)^2}$, but you would have to give a different name to it because everyone else has already agreed that "absolute value" means to take $\sqrt{a^2+b^2}$ instead.
Now, a more interesting question is why everyone else decided on that definition. One reason is that you can represent complex numbers as points in the plane by letting $a+bi$ correspond to the point $(a,b)$, and then $|a+bi|$ is the distance from this point to the origin. Note that when you do this, $(a,b)$ is just a point in the ordinary Euclidean plane: it is a point that is $a$ units to the right of the origin and $b$ units above the origin. It doesn't make sense to say that the vertical distance is $bi$, since in geometry when we measure distances they are always positive real numbers. The vertical distance is $b$ because you have moved $b$ units vertically in the plane. (Actually, this is only accurate if $b$ is positive: if $b$ is negative, you have moved $-b$ units down, and so the distance is $-b$ rather than $b$. But you end up squaring this quantity when you use the Pythagorean theorem, so it doesn't matter if it's negative.)
Now ultimately this explanation is not very satisfying, because it doesn't explain why we chose to represent complex numbers in the plane this way. For instance, why don't we choose to represent them such that the complex number $i$ corresponds to a vertical distance different from $1$, or a distance in some direction other than vertical? One answer is that choosing $i$ to mean "go up one unit" happens to make distance have nice algebraic properties we would like absolute values to have. For instance, for real numbers, it is true that $|xy|=|x||y|$. Defining absolute values of complex numbers by $|a+bi|=\sqrt{a^2+b^2}$, it turns out that this is true for complex numbers as well. As a simple example, if we want $|xy|=|x||y|$ to be true for complex numbers, then we should have $|i|^2=|i^2|=|{-1}|=1$. So we should define $|i|=1$ or $-1$, and it is sensible to define it to be $1$ instead of $-1$ since absolute values are supposed to be positive.
Best Answer
Using the triangle inequality $$|z^2|-|2z\cos(\alpha)|\leq |z^2+2z\cos(\alpha)|\leq 1$$ but $$|z^2|-|2z|\leq |z^2|-|2z\cos(\alpha)|$$ therefore $$|z^2|-|2z|=|z|^2-2|z|\leq 1\Rightarrow (|z|-1)^2\leq2\Rightarrow ||z|-1|\leq\sqrt{2}$$ In other words $$0 \leq |z|\leq \sqrt{2}+1$$ Notice that this inequality is sharp because it is attained for $\alpha=\pi$ and $z=\sqrt{2}+1$.