Question

Numere complexe!
Exercițiile din poza!

Mă mulțumesc și doar cu unul​

Answer 1

Răspuns:

Explicație pas cu pas:

la 3 deja ai primit raspuns, acum iti raspund la 4.

$z _1^2 + z_2^2 + z_3^2 = 0\\\\z _1^2 = - z_2^2 - z_3^2\\\\z _2^2 = - z_1^2 - z_3^2\\\\\\z _3^2 = - z_1^2 - z_2^2$

Fie numarul A , astfel incat

$\Big A = \frac{z_1^6 + z_2^6 + z_3^6}{z_1^2\cdot z_2^2\cdot z_3^2} =$

$= \frac{z_1^6 }{z_1^2\cdot z_2^2\cdot z_3^2} + \frac{z_2^6 }{z_1^2\cdot z_2^2\cdot z_3^2} + \frac{z_1^6 }{z_3^2\cdot z_2^2\cdot z_3^2} =$

$= \frac{z_1^4 }{z_2^2\cdot z_3^2} + \frac{z_2^4 }{z_1^2\cdot z_3^2} + \frac{z_3^4 }{z_3^2\cdot z_2^2} =$

$=\Big( \frac{z_1^2 }{z_2\cdot z_3}\Big)^2 + \Big( \frac{z_2^2 }{z_1\cdot z_3}\Big)^2 + \Big( \frac{z_3^2 }{z_1\cdot z_2}\Big)^2 =$

$=\Big( \frac{-z_2^2 -z_3^2 }{z_2\cdot z_3}\Big)^2 + \Big( \frac{-z_1^2 -z_3^2 }{z_1\cdot z_3}\Big)^2 + \Big( \frac{-z_1^2 -z_2^2 }{z_1\cdot z_2}\Big)^2 =$

$=\Big( \frac{z_2^2+z_3^2 }{z_2\cdot z_3}\Big)^2 + \Big( \frac{z_1^2 +z_3^2 }{z_1\cdot z_3}\Big)^2 + \Big( \frac{z_1^2 +z_2^2 }{z_1\cdot z_2}\Big)^2 =$

$=\Big( \frac{z_2^2}{z_2\cdot z_3}+\frac{z_3^2}{z_2\cdot z_3}\Big)^2 + \Big( \frac{z_1^2}{z_1\cdot z_3}\Big+\frac{z_3^2}{z_1\cdot z_3}\Big)^2 + \Big( \frac{z_1^2}{z_1\cdot z_2}+\frac{z_2^2}{z_1\cdot z_2}\Big)^2 =$

$=\Big( \frac{z_2}{z_3}+\frac{z_3}{z_2}\Big)^2 + \Big( \frac{z_1}{z_3}\Big+\frac{z_3}{z_1}\Big)^2 + \Big( \frac{z_1}{z_2}+\frac{z_2}{z_1}\Big)^2 =$

$=\Big( \frac{z_2}{z_3}\Big)^2 + 2* \frac{z_2}{z_3}* \frac{z_3}{z_2} +\Big( \frac{z_3}{z_2}\Big)^2 + \Big( \frac{z_1}{z_3}\Big)^2 + 2* \frac{z_1}{z_3}* \frac{z_3}{z_1} +\Big( \frac{z_3}{z_1}\Big)^2 + \Big( \frac{z_1}{z_2}\Big)^2 + 2* \frac{z_1}{z_2}* \frac{z_2}{z_1} +\Big( \frac{z_2}{z_1}\Big)^2 =$

$=\frac{z_2^2}{z_3^2} + 2 +\frac{z_3^2}{z_2^2} + \frac{z_1^2}{z_3^2} + 2 +\frac{z_3^2}{z_1^2} +\frac{z_1^2}{z_2^2} + 2 +\frac{z_2^2}{z_1^2} =$

$=\frac{z_1^2+ z_2^2}{z_3^2} + \frac{z_2^2+ z_3^2}{z_1^2} + \frac{z_1^2+ z_3^2}{z_2^2} + 6=$

$=\frac{-z_3^2}{z_3^2} +\frac{-z_1^2}{z_1^2} +\frac{-z_2^2}{z_2^2} +6 =$

$=-1 -1 -1 + 6 = 3$

deci

$\frac{z_1^6 + z_2^6 + z_3^6}{3z_1^2\cdot z_2^2\cdot z_3^2} =\frac{A}{3} =\frac{3}{3} = 1$