Question

sa se determine z inclus in C stiind ca a) z²=(1-i)/(1+i)
b)z²=(-2+4i)/(2+i)
c)z conjugat=z²​

Answer 1

Explicație pas cu pas:

a)

z = x + yi, unde x,y∈ℝ <=> z²=x²+2xyi-y²

$\dfrac{1 - i}{1 + i} = \dfrac{ {(1 - i)}^{2} }{(1 + i)(1 - i)} = \dfrac{1 - 2i - 1}{1 + 1} = - \dfrac{2i}{2} = - i$

$\implies {x}^{2} + 2xyi - {y}^{2} = - i \\ ({x}^{2} - {y}^{2}) + (2xy + 1)i = 0$

$\begin{cases}{x}^{2} - {y}^{2} = 0 \\2xy + 1 = 0 \end{cases} \iff \begin{cases}(x - y)(x + y) = 0 \\x = - \dfrac{1}{2y} \end{cases}$

$(- \dfrac{1}{2y} - y)(- \dfrac{1}{2y} + y) = 0$

$- \dfrac{1}{2y} - y = 0 \iff 2 {y}^{2} + 1 = 0 \to x \not \in \mathbb \: {R}$

$- \dfrac{1}{2y} + y = 0 \iff 2 {y}^{2} - 1 = 0$

${y}^{2} = \dfrac{1}{2} \implies y = \pm \dfrac{ \sqrt{2} }{2}\implies x = \mp \dfrac{ \sqrt{2} }{2}$

$\implies \begin{cases}z = \ \ \ \dfrac{ \sqrt{2} }{2} - \dfrac{ \sqrt{2} }{2}i \\z = - \dfrac{ \sqrt{2} }{2} + \dfrac{ \sqrt{2} }{2}i \end{cases}$