Question

ECUACIONES CON NÚMERO COMPLEJOS

Resolver:

a) Z + 2i = 2 Z -7



b) 3 Z + 2i = 2 Z – 5



c) 7i + Zi = 2 Z – 11



d) Zi + 9i = -15 + 2 Z

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Answer 1

Explicación paso a paso:

a) Z + 2i = 2 Z -7

z - 2z = -7 - 2i

-z = (-7-2i):2

-z = -7/2 - i

z = 7/2 + i

b) 3 Z + 2i = 2 Z – 5

3z - 2z = -5 - 2i

z = -5 - 2i

c) 7i + Zi = 2 Z – 11

zi - 2z = -11 - 7i

z.(i-2) = -11 - 7i

$z = \frac{ - 11 - 7i}{ - 2 + i} \times \frac{ - 2 - i}{ - 2 - i} =$

$z = \frac{22 + 11i + 14i + 7i {}^{2} }{( - 2) {}^{2} - (i) {}^{2} }$

$z = \frac{22 + 11i + 14i + 7.( - 1)}{4 - i {}^{2} }$

$z = \frac{22 + 25i - 7}{4 - ( - 1)}$

$z = \frac{15 + 25i}{4 + 1}$

$z = \frac{15 + 25i}{5}$

$z = 3 + 5i$

d) Zi + 9i = -15 + 2 Z

zi - 2z = -15-9i

z.(i -2) = -15-9i

$z= \frac{ - 15 - 9i}{ - 2 + i} \times \frac{ - 2 - i}{ - 2 - i}$

$z = \frac{30 + 15i + 18i + 9i {}^{2} }{( - 2) {}^{2} - i {}^{2} }$

$z = \frac{30 + 33i - 9}{4 + 1}$

$z = \frac{21 + 33i}{5}$

$z = \frac{21}{5} + \frac{33}{5} i$