The system of equations

Question

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:

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

Answer 1

First, let $\gamma=z^2+w^2$. Now note that \begin{align} a^2+2(2b)^2+2(2c)^2&=(x^2+y^2)^2+\gamma^2\\ &=(a+\gamma)^2+\gamma^2\\ 8b^2+8c^2&=2a\gamma+2\gamma^2\\ 16b^2+16c^2&=4a\gamma+4\gamma^2 \end{align} Now note that $2\gamma=1-a$ and $b^2+c^2=1-a^2$, so that this rewrites to $$16-16a^2=2a(1-a)+(1-a)^2$$ which we can solve for $a$, obtaining $a=1$ or $a=-1$. In both cases, $a^2+b^2+c^2=1$ rewrites to $b^2+c^2=0$, and since squares are non-negative, we can conclude that $b=c=0$.

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So we need to solve (for $a=1$ and $a=-1$): \begin{align} x^{2}+y^{2}+z^{2}+w^{2}&=1\\ x^{2}+y^{2}-z^{2}-w^{2}&=a\\ xw+yz&=0\\ yw-xz&=0 \end{align}

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Case one: $a=1$

We immediately see that $1-a=z^2+w^2=0$, so that $z=w=0$. We are left with $x^2+y^2=0$, and see that the general solution is given by $$(x,y,z,w)=(\cos(\theta),\sin(\theta),0,0)$$ for any real $\theta$.

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Case two: $a=-1$

From this, we see $a+1=x^2+y^2=0$, and so $x=y=0$; and we are left with $z^2+w^2=1$. Now the general solution becomes

$$(x,y,z,w)=(0,0,\cos(\theta),\sin(\theta))$$ for any real theta.

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Therefore,

All solutions are given by, for any $\theta\in\Bbb R$: $$(x,y,z,w)=(\cos(\theta),\sin(\theta),0,0)$$ or $$(x,y,z,w)=(0,0,\cos(\theta),\sin(\theta))$$

Answer 2

Hint (assuming $x,y,z,w$ are reals): let $u=y+ix$ and $v=w+iz$ then the system becomes:

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

From the first 2 equations $|u|^2 = \frac{1}{2}(1+a)\,$, $|v|^2=\frac{1}{2}(1-a)\,$, so $|uv|^2 =\frac{1}{4}(1-a^2)=\frac{1}{4}(b^2+c^2)$. From the third equation $|uv|^2=4(c^2+b^2)\,$, and equating the two gives $b=c=0 \implies uv=0$. The rest should be straightforward to fill in.