Appointed-at-the-nodal-level-of-the-group-z-z-2-z-z-4-

Question Number 127366 by mohammad17 last updated on 29/Dec/20

A p p o i n t e d a t t h e n o d a l l e v e l o f t h e g r o u p

∣ z + \bar{z} ∣^{2} + (z - \bar{z}) = 4 ?

Commented by JDamian last updated on 29/Dec/20

z = \pm 1

Commented by mohammad17 last updated on 29/Dec/20

h o w s i r c a n y o u g i v e m e s t e p s ?

Commented by JDamian last updated on 29/Dec/20

z \equiv a + b i

z + \bar{z} = 2 a

z - \bar{z} = 2 b i

∣ z + \bar{z} ∣^{2} + (z - \bar{z}) = 4

∣ 2 a ∣^{2} + 2 b i = 4 \Rightarrow {\begin{cases} b = 0 \\ 4 ∣ a ∣^{2} = 4 \Rightarrow∣ a ∣^{2} = 1 \Rightarrow a = \pm 1 \end{cases}

Commented by mohammad17 last updated on 29/Dec/20

t h a n k y o u s i r a r e y o u c a n h e l p m e i n a l l a t e r q u e s t i o n ?

Answered by liberty last updated on 29/Dec/20

z = a + b i \Rightarrow \bar{z} = a - b i

∣ z + \bar{z} ∣^{2} = 4 a^{2}; z - \bar{z} = 2 b i

\Leftrightarrow 4 a^{2} + 2 b i = 4 + 0 i \to {\begin{cases} 4 a^{2} = 4 \Rightarrow a = \pm 1 \\ b = 0 \end{cases}

∴ z = \pm 1

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