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For fixed complex number $s≠0$, what 4-dim shape is given by complex solutions $z,w$ of $z^2+w^2=s^2$ and higher dimensional version 2N-dim shapes of $z_1^2+z_2^2+...+z_N^2=s^2$ ?

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Write $z^2 + w^2 = (z + iw)(z - iw) = s^2$ and write $u = z + iw, v = z - iw$. This is an invertible change of coordinates since we have $z = \frac{u + v}{2}, w = \frac{u - v}{2i}$, so it suffices to determine the shape of $uv = s^2$ for fixed $s^2 \neq 0$. But given an arbitrary $u \neq 0 \in \mathbb{C}$ we have $v = \frac{s^2}{u}$ is uniquely determined, so our mystery shape is just $\mathbb{C} \setminus \{ 0 \}$.

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So the subset of $\mathbb{C}^2$ that satisfies the equation is $\mathbb{C}^2 \setminus \{(0,0)\}$ – vtt May 31 '12 at 17:46
@vtt: no. It consists of all pairs of the form $\left( \frac{u + \frac{s^2}{u}}{2}, \frac{u - \frac{s^2}{u}}{2i} \right)$ for $u \in \mathbb{C} \setminus \{ 0 \}$. – Qiaochu Yuan May 31 '12 at 18:01

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