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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?

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  • $\begingroup$ If $z$ and $w$ satisfy the conditions then so do $ze^{i\alpha}$ and $we^{i\alpha}$ for any $\alpha\in\mathbb{R}$. So you can just choose for e.g. $z=\left|z\right|\in\left[0,\infty\right)$. $\endgroup$ – drhab Jan 10 '14 at 0:25
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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 $z=9$, $w=-7$.

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(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:

$\left(2r-a\right)^{2}+b^{2}=625$

$\left(r+2a\right)^{2}+4b^{2}=25$

$\left(r+a\right)^{2}+b^{2}=4$

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Hint: Think geometrically. Draw a picture. You will have some parallelograms and using some trigonometry you can deduce the answer.

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