Complex numbers - system of equations Let $z$ and $w$ be complex numbers such that $|2z - w| = 25$, $|z + 2w| = 5$, and $|z + w| = 2$. Find $|z|$.
Any tips where to start? Is there a better way then just squaring both sides and solving then replugging them in?
 A: The given conditions are equivalent with the equations
$$\eqalign{
4 z\bar z-2 z\bar w- 2\bar z w+w\bar w&=625\cr
z\bar z+2z\bar w+2\bar z w+4w\bar w&=25\cr
z\bar z+z\bar w+\bar z w +w\bar w&=4\cr}\tag{1}$$
Multiplying them with $1$, $1$,  $-5$ respectively, and adding leads to
$$z\bar w+\bar z w=-126\ ,\tag{2}$$
so that we arrive at
$$4z\bar z+w\bar w =373,\qquad z\bar z+4w\bar w=277\ ,$$
which immediately implies $|z|=9$, $\>|w|=7$, or
$$z=9e^{i\alpha}, \quad w=7 e^{i\beta}\ .$$
Plugging this into $(2)$ we conclude that necessarily $\cos(\alpha-\beta)=-1$, or $\beta=\alpha+\pi$ $\>(2\pi)$. It follows that the solutions of $(1)$ are necessarily of the form
$$z=9e^{i\alpha},\quad w=-7e^{i\alpha}\qquad(\alpha\in{\mathbb R})\ .\tag{3}$$
We now have to check whether the pairs $(3)$ do indeed fulfill $(1)$. Since the left sides of $(1)$ are obviously independent of $\alpha$ it is enough to check with $z=9$, $w=-7$.
A: (not a full answer, but more a tip)
Assuming that $z=r\in\left[0,\infty\right)$ (see my comment) and
$w=a+bi$ you must solve $r$ from:
$\left(2r-a\right)^{2}+b^{2}=625$
$\left(r+2a\right)^{2}+4b^{2}=25$
$\left(r+a\right)^{2}+b^{2}=4$
A: Hint: Think geometrically. Draw a picture. You will have some parallelograms and using some trigonometry you can deduce the answer. 
