Solve-in-C-x-y-z-1-x-1-y-1-z-

Question Number 177702 by lapache last updated on 08/Oct/22

S o l v e i n C

{\begin{cases} x + y = z \\ \frac{1}{x} + \frac{1}{y} = \frac{1}{z} \end{cases}

Answered by Frix last updated on 08/Oct/22

(1) x = z - y

(2) x = \frac{y z}{y - z}

\Leftrightarrow

z - y = \frac{y z}{y - z}

y^{2} + y z - z^{2} = 0 \Rightarrow y = z \times (\frac{1}{2} \pm \frac{\sqrt{3}}{2} i)

\Rightarrow

x = z \times (\frac{1}{2} \mp \frac{\sqrt{3}}{2} i)

z \in C \Rightarrow z = p + q i

(\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} (p + q i) (\frac{1}{2} \pm \frac{\sqrt{3}}{2} i) \\ (p + q i) (\frac{1}{2} \mp \frac{\sqrt{3}}{2} i) \\ p + q i \end{matrix})

Commented by Tawa11 last updated on 08/Oct/22

Great sir

Answered by mr W last updated on 08/Oct/22

\frac{x + y}{x y} = \frac{1}{z}

\Rightarrow x y = (x + y) z = z^{2}

x + y = z

\Rightarrow x, y a r e r o o t s o f t^{2} - z t + z^{2} = 0

x, y = \frac{z \pm \sqrt{z^{2} - 4 z^{2}}}{2} = \frac{(1 \pm 3 i) z}{2}

Commented by Tawa11 last updated on 08/Oct/22

Great sir

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