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Let $a,b,c\in \mathbb{R}$ that $a^{2}+b^{2}+c^{2}=1$. We want to find $x,y,z,w$ in the following equations:

$$\begin{cases} \begin{align} x^{2}+y^{2}&=\frac{1}{2}(1+a) \tag{1}\\ w^{2}+z^{2}&= \frac{1}{2}(1-a) \tag{2}\\ xw+yz&= \frac{1}{2} b \tag{3}\\ yw-xz&= \frac{1}{2} c.\tag{4} \end{align} \end{cases}$$

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  • $\begingroup$ and what follows after the plus sign? $\endgroup$ – Dr. Sonnhard Graubner Mar 31 '17 at 18:27
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    $\begingroup$ I assume that the + was a typo for &, and edited accordingly. $\endgroup$ – Théophile Mar 31 '17 at 18:31
  • $\begingroup$ Yes. Thanks so much $\endgroup$ – 444 Mar 31 '17 at 18:34
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Hint (assuming $x,y,z,w$ are reals): let $u=y+ix$ and $v=w+iz$ then the system becomes:

$$ \begin{cases} \begin{align} |u|^2 &= \frac{1}{2}(1+a) \\ |v|^2 &= \frac{1}{2}(1-a) \\ uv & = \frac{1}{2}(c+ib) \end{align} \end{cases} $$

From the first two equations $|u|^2|v|^2= \frac{1}{4}(1-a^2)=\frac{1}{4}(b^2+c^2)\,$. From the third equation $|uv|^2=\frac{1}{4}(c^2+b^2)\,$ as well, so the modulus part of the 3rd equation is redundant. Then pick $u$ to be any complex number of modulus $\sqrt{\frac{1+a}{2}}\,$, and let $v=\frac{c+ib}{2u}$.

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