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Suppose $z$ is a complex number. Prove that there exists an $r \geq 0$ and a complex number $w$ (with $|w| = 1$) such that $z = rw$. Does $z$ uniquely determine $r$ and $w$?

Let $z = a+bi$. Then $|z| = \sqrt{a^2+b^2}$. So take $r = |z|$. It seems like one can take $w$ to be any complex number such that $|w| = 1$. So I think $z$ uniquely determines $r$ but not $w$.

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    This is the polar representation of a complex number.2010-12-21

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Let $k\in\mathbb{R}, z\in\mathbb{C}, z=a+bi$. Then $kz = ka+kbi$. Therefore $w$ is uniquely determined by $z$, and $w = \frac{a}{r} + \frac{b}{r}i$.

You can also think of this using vectors in $\mathbb{R}^2$; any vector can be represented as the unit vector with its same angle multiplied by its length.

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    There is uniqueness except for $z=0$.2010-12-21
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    Indeed. $z=0$ determines $r=0$ but not $w$.2010-12-21
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No, if you take $w=1+0i$ for example, $wr$ will always be real. So $w$ must have the same phase as $z$.

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    I see. I guess it depends on on defines $|z|$.2010-12-22
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    @Trevor: it does not depend upon |z|. arg (w) must be the same as arg(z), as arg (|z|)=0.2010-12-22