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Ecuatie cu coeficienti complecsi
Posted: Tue Jan 26, 2010 10:58 pm
by alex2008
Sa se arate ca daca radacinile ecuatiei cu coeficienti complecsi \( az^2+bz+c=0 \) au acelasi modul, atunci \( |a|\cdot \overline{b}\cdot c=\overline{a}\cdot b\cdot |c| \).
Posted: Tue Jan 26, 2010 11:31 pm
by Mateescu Constantin
\( |z_1|=|z_2|\ \Longleftrightarrow\ z_1\overline{z_1}=z_2\overline{z_2}\ \Longleftrightarrow\ z_1\overline{z_1}+\overline{z_1}z_2=z_2\overline{z_2}+\overline{z_1}{z_2}\ \Longleftrightarrow\ \overline{z_1}(z_1+z_2)=z_2(\overline{z_1+z_2}) \\\ \\\
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\Longleftrightarrow\ z_1\overline{z_1}(z_1+z_2)=z_1z_2(\overline{z_1+z_2})\ \Longleftrightarrow\ |z_1|^2(z_1+z_2)=z_1z_2(\overline{z_1+z_2})\\\ \\\
\\\ \\\
\Longleftrightarrow^{\normal |z_1|=|z_2|}\ |z_1z_2|(z_1+z_2)=z_1z_2(\overline{z_1+z_2})\ \Longleftrightarrow\ \left|\frac ca\right|\ \cdot\ \left\(-\frac ba\right\)=\frac ca\ \cdot\ \left\(-\frac{\overline b}{\overline a}\right\)\ \Longleftrightarrow\ \overline{\underline{\left\|\ \overline ab|c|=|a|\overline bc\ \right\|}} \)
Posted: Tue Jan 26, 2010 11:38 pm
by Marius Mainea
Folosind relatiile lui Viete, concluzia este echivalenta cu \( \frac{\overline{b}}{\overline{a}}\cdot\frac{c}{a}\cdot\frac{a}{b}=|\frac{c}{a}| \) sau \( |z_1z_2|=\overline{z_1+z_2}\cdot (z_1z_2)\frac{1}{z_1+z_2} \) sau \( (z_1+z_2)|z_1|^2=|z_1|^2z_2+|z_2|^2z_1 \) care este adevarata avand in vedere ipoteza.