Exercise 1.11

Question

Exercise 11: If $z \in \mathbb C$, prove that there exists an $ r \geq 0$ and $w \in \mathbb C$ subh that $z = rw$.
    Are $w$ and $r$ always uniquely determined by $z$?

ghostofgarborg · Accepted Answer

There is a solution, and it is unique whenever $z \neq 0$. Assume $z = rw$, $r \geq 0$ and $|w|^2 = w \bar w = 1$.
    $$ r = \sqrt{r^2} = \sqrt{(rw)(\overline{rw})} = \sqrt{z \bar z} = |z| $$
    If $r = 0$, we can take any $w$ with $|w| = 1$, e.g. $w = e^{i \theta}$. Otherwise
    $$ w = \frac{z}{r} = \frac{z}{|z|} $$
    It is easy to check that the uniquely determined $r$ and $w$ have the desired properties.

Amit Awasthi · Answer

\begin{proof}
      Let $z = a + bi$, and let
      \begin{equation*}
        r = |z| = \sqrt{a^2 + b^2} \ge 0.
      \end{equation*}
      Let $w = a/r + bi/r$ when $r > 0$. Then,
      \begin{equation*}
        |w| = \sqrt{\left(\frac{a}{r}\right)^2+\left(\frac{b}{r}\right)^2} = \frac{\sqrt{a^2+b^2}}{r} = 1,
      \end{equation*}
      and $z = rw$ by construction. $r$ is uniquely determined since $|z| = |rw| = r$ since $r \ge 0$. $w$ is uniquely determined when $|z| = r > 0$ by construction. When $|z| = r = 0$, however, $w$ is not uniquely determined since we can choose $w$ to be any complex number such that $|w| = 1$ since $z = 0 = 0w = rw$ for all such $w$.
\end{proof}

Exercise 1.11

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Exercise 1.11

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