Finding range of $||x| - |y||$ for the given conditions. 
If $ z = x + iy$ and $ x^2 + y^2 = 16 $ , then the range of $||x|-|y||$ is...? 

This is what I've tried yet: 
Suppose $x = a\cos \theta$ and $y = b\sin \theta$,  then we've :
$$\begin{align}
x^2 + y^2 = 16 \implies & a^2\cos^2 \theta + b^2\sin^2 \theta= 16 \\
\implies & a^2 + \sin^2 \theta(b^2 - a^2) = 16 \\
\implies & \sin^2 \theta = \cfrac{16 - a^2}{b^2 - a^2} \tag{1}
\end{align}$$
Not sure from where to go on from here. Putting the same values in $ z = x+iy$ and then squaring both sides doesn't give anything fruitful too. 
 A: Hint:-
Put $x=4\sin\theta$ and $y=4\cos \theta$. 
Solution

 Then, $$||x|-|y||=4||\sin\theta|-|\cos \theta||$$Now, $$0\le|\sin\theta|\le 1$$and, $$-1\le-|\cos\theta|\le0$$which gives, $$||\sin\theta|-|\cos \theta||\le1\implies 4||\sin\theta|-|\cos \theta||\le4$$ 

A: If $x, y \in \mathbb{R} $, then you find that the range is $[0,4]$. In fact, just because $||x|-|y|| \geq 0$ and you can easily find $x, y$ such that $x=y \wedge x^2+y^2=16$. The other bound is given by this argument: you can assume $x, y \geq 0$, in fact if they are both $\leq 0$ you get the same results multiplying by $-1$, and if $x\geq 0 \wedge y<0$ than you apply the same argument.
So, since they are both $\geq 0$, assuming $x\leq y$, you get $y \leq 4$, in other cases $x^2+y^2 \geq y^2 > 16$. So $||y|-|x|| \leq 4$.
A: Taking $x=a\cos\theta$ and $y=a\sin\theta$ won't be helpful because all you will be doing is introducing two more variables a and b making the solution more messy.. 
So in such cases think about a case to simplify either RHS or LHS..
Here this can be done by assuming $x=4\sin\theta$ and $y=4\cos\theta$
In most of these questions, keeping the objective of question in mind I'd helpful... 
