Question

Acest ex.Vă rog . ..​

Answer 1

Explicație pas cu pas:

$z^{2} = i \overline{z}$

$z = a + b i$

$\overline{z} = a - bi$

=>

${(a + b i)}^{2} = i(a - b i)$

$( {a}^{2} - {b}^{2} ) + 2ab \ i = b + a \ i$

=>

$\begin{cases} {a}^{2} - {b}^{2} = b\\ 2ab = a\end{cases}$

dacă:

$a = 0 \implies b(b + 1) = 0$

$\bf \implies \begin{cases} a = 0; b = -1\\a = 0; b = 0 \end{cases}$

dacă:

$a \not= 0 \implies 2b = 1 \implies b = \frac{1}{2}$

${a}^{2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$

$\bf \implies \begin{cases} a = - \frac{ \sqrt{3} }{2} ; b = \frac{1}{2} \\a = \frac{ \sqrt{3} }{2} ; b = \frac{1}{2} \end{cases}$