求解 {l}{x^2+y^2=8}{z=x+y}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { x ^ {2} + y ^ {2} = 8 }\\ { z = x + y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

求解 x, y, z 的值 (复数求解)

$$\left\{\begin{matrix}x=\frac{\sqrt{-2b\sqrt{16-b^{2}}+16}}{2}\text{, }y=\frac{\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&a=b\text{ and }arg(\frac{-\sqrt{16-b^{2}}+b}{2})<\pi \\x=\frac{\sqrt{2b\sqrt{16-b^{2}}+16}}{2}\text{, }y=\frac{-\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&\left(a=-2\sqrt{2}\text{ and }b=-2\sqrt{2}\right)\text{ or }\left(a=b\text{ and }arg(\frac{\sqrt{16-b^{2}}+b}{2})<\pi \right)\\x=-\frac{\sqrt{-2b\sqrt{16-b^{2}}+16}}{2}\text{, }y=\frac{\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&a=b\text{ and }arg(\frac{\sqrt{16-b^{2}}-b}{2})<\pi \\x=-\frac{\sqrt{2b\sqrt{16-b^{2}}+16}}{2}\text{, }y=\frac{-\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&\left(a=-2\sqrt{2}\text{ and }b=-2\sqrt{2}\right)\text{ or }\left(a=b\text{ and }arg(\frac{-\sqrt{16-b^{2}}-b}{2})<\pi \right)\\x=0\text{, }y=2\sqrt{2}\approx 2.828427125\text{, }z=2\sqrt{2}\approx 2.828427125\text{, }&a=2\sqrt{2}\text{ and }b=2\sqrt{2}\end{matrix}\right.$$

求解 x, y, z 的值

$$\left\{\begin{matrix}x=-\frac{\sqrt{2\left(-b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&a=b\text{ and }b\leq 2\sqrt{2}\text{ and }|b|\leq 4\\x=-\frac{\sqrt{2\left(b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{-\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&a=b\text{ and }b\geq -4\text{ and }b\leq -2\sqrt{2}\\x=\frac{\sqrt{2\left(-b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&a=b\text{ and }b\geq 2\sqrt{2}\text{ and }b\leq 4\\x=\frac{\sqrt{2\left(b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{-\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&a=b\text{ and }b\geq -2\sqrt{2}\text{ and }|b|\leq 4\end{matrix}\right.$$