let x,y,z be a complex numbers as ∣x∣=∣y∣=∣z∣=1 { ((x+y+z=1)),((xyz=1)) :} calcul (1/x)+(1/y)+(1/z)=? x=? y=? z=? please i need a help


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Question Number 107320 by pticantor last updated on 10/Aug/20

$l e t x, y, z b e a c o m p l e x n u m b e r s$ $a s ∣ x ∣=∣ y ∣=∣ z ∣= 1$ ${\begin{cases} x + y + z = 1 \\ x y z = 1 \end{cases}$ $c a l c u l \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = ?$ $x = ? y = ? z = ?$ $p l e a s e i n e e d a h e l p$
Commented by bemath last updated on 10/Aug/20

${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + x z + y z)$ $x^{2} + y^{2} + z^{2} = 1 - 2 (x y + x z + y z)$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{x + y}{x y} + \frac{1}{z} = \frac{x z + y z + x y}{x y z}$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = x y + x z + y z$ $= \frac{1 - (x^{2} + y^{2} + z^{2})}{2}$ $n e e d m o r e i n f o r m a t i o n$
Commented by pticantor last updated on 10/Aug/20

$y e s s i r ∣ x ∣=∣ y ∣=∣ z ∣= 1$
Commented by bemath last updated on 10/Aug/20

$i f ∣ x ∣=∣ y ∣=∣ z ∣= 1$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1 - 3}{2} = - 1$
Commented by pticantor last updated on 10/Aug/20

$h o w s i r ? ? ?$ $You can't use 'macro parameter character #' in math mode$
Commented by bemath last updated on 10/Aug/20
$if ∣x∣ = 1 it does mean → { ((x=−1 or)),((x=1)) :} or x^2 =1$
$i f ∣ x ∣ = 1 i t d o e s m e a n \to {\begin{cases} x = - 1 o r \\ x = 1 \end{cases}$ $o r x^{2} = 1$
Commented by Her_Majesty last updated on 10/Aug/20

$∣ x ∣ \Rightarrow x = c o s θ + i s i n θ = e^{i θ}$
	Answered by Her_Majesty last updated on 10/Aug/20

	$e^{i (α + β + γ)} = 1 \Rightarrow α + β + γ = 0 \Leftrightarrow γ = - (α + β)$ $e^{i α} + e^{i β} + e^{i γ} = 1$ $\Rightarrow$ $\cos α + c o s β + c o s (α + β) = 1$ $s i n α + s i n β - s i n (α + β) = 0$ $\Rightarrow i f α < β < γ : α = - π / 2 β = 0 γ = π / 2$ $\Rightarrow x = - i y = 1 z = i$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$
	Commented by bemath last updated on 10/Aug/20

	$h o w ∣ x ∣= 1 \Rightarrow x = - i . i m p o s s i b l e$ $∣ x ∣^{2} = 1^{2} \Rightarrow x^{2} = 1$ $b u t i f x = - i \Rightarrow x^{2} = i^{2} = - 1$
	Commented by Her_Majesty last updated on 10/Aug/20

	${y o u}^{'} r e w r o n g$ $∣ a + b i ∣= \sqrt{a^{2} + b^{2}}$ $∣ 0 \pm 1 i ∣= \sqrt{0^{2} + 1^{2}} = 1$
	Commented by Her_Majesty last updated on 10/Aug/20

	$w h o m a r k e d t h i s a s i n a p p r o p r i a t e ?$ $t h e q u e s t i o n s a y s ‘ ‘ x y z a r e c o m p l e x {n u m b e r s}^{″}$ $z = {r e}^{i θ} = r c o s θ + i r s i n θ = a + b i w i t h r \in R^{+}$ $∣ z ∣= r = \sqrt{a^{2} + b^{2}}$ $i f y o u d o n o t k n o w t h i s, l e a r n i t r i g h t n o w!$
	Commented by bemath last updated on 10/Aug/20
	wow .... I don't know who did that action.
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