[tex]z^2=\frac{1-i}{1+i}=\frac{(1-i)^2}{2}=\frac{-2i}{2}=-i\\
z^2=-i\\
Putem\ rezolva\ in\ doua\ moduri:\\
i)luam\ z=a+bi,a,b\in R\ si\ facem\ calcule.\\
ii)folosim\ forma\ trigonometrica\ a\ numarului\ complex.\\
Le\ luam\ pe\ amandoua\ la\ rand:\\
i)z=a+bi,a,b\in R\\
(a+bi)^2=-i\\
a^2+2abi-b^2=-i\\
\left \{ {{a^2-b^2=0} \atop {2ab=-1}} \right.\\
Din\ prima\ ecuatie\ avem:a=+\_b\\
Daca\ a=b:\\
2b^2=-1\\
b^2=-\frac{1}{2} (nu\ convine)\\
Daca\ a=-b:\\
-2b^2=-1\\
[/tex]
[tex]b^2=\frac{1}{2}\Rightarrow b\in \{\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\}\\
Asadar\ z=\frac{1}{\sqrt2}-\frac{i}{\sqrt{2}}\ sau\ z=-\frac{1}{\sqrt2}+\frac{i}{\sqrt2}\\
\\
ii)Folosim\ forma\ trigonometrica\ a\ numarului\ complex:\\
r=|-i|=1\\
\cos\Phi=0\\
\sin \Phi=1\Rightarrow \Phi=\frac{\Pi}{2}\\
z_k=\cos\frac{\Phi+2k\Pi}{2}+i\cdot \sin \frac{\Phi+2k\Pi}{2},k=\overline{0,1}\\
z_0=\cos \frac{\Pi}{4}+i\codt \sin \frac{\Pi}{4}\\
z_1=\cos \frac{5\Pi}{4}+i\cdot \sin \frac{5\Pi}{4}\\
Sper\ ca\ te-am\ ajutat.[/tex]
