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For z=x+iy𝑧=𝑥+𝑖𝑦 find the values x𝑥 and y𝑦 satisfying 2z1+i−2zi=52+i2𝑧1+𝑖−2𝑧𝑖=52+𝑖.

Question

For z=x+iy𝑧=𝑥+𝑖𝑦 find the values x𝑥 and y𝑦 satisfying 2z1+i−2zi=52+i2𝑧1+𝑖−2𝑧𝑖=52+𝑖.

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Solution

Para resolver la ecuación 2z(1+i)2z(i)=5(2+i)2z(1+i) - 2z(i) = 5(2+i), primero descomponemos zz en sus partes real e imaginaria, es decir, z=x+iyz = x + iy.

  1. Sustituimos zz en la ecuación: 2(x+iy)(1+i)2(x+iy)i=5(2+i) 2(x + iy)(1 + i) - 2(x + iy)i = 5(2 + i)

  2. Expandimos los términos: 2(x+iy)(1+i)=2x(1+i)+2iy(1+i)=2x+2xi+2iy+2i2y=2x+2xi+2iy2y 2(x + iy)(1 + i) = 2x(1 + i) + 2iy(1 + i) = 2x + 2xi + 2iy + 2i^2y = 2x + 2xi + 2iy - 2y 2(x+iy)i=2xi2i2y=2xi+2y -2(x + iy)i = -2xi - 2i^2y = -2xi + 2y

  3. Sumamos los términos: 2x+2xi+2iy2y2xi+2y=2x+2iy 2x + 2xi + 2iy - 2y - 2xi + 2y = 2x + 2iy

  4. Igualamos al lado derecho de la ecuación: 2x+2iy=5(2+i)=10+5i 2x + 2iy = 5(2 + i) = 10 + 5i

  5. Igualamos las partes reales e imaginarias: 2x=10y2y=5 2x = 10 \quad \text{y} \quad 2y = 5

  6. Resolviendo para xx y yy: x=102=5 x = \frac{10}{2} = 5 y=52=2.5 y = \frac{5}{2} = 2.5

Por lo tanto, los valores de xx y yy que satisfacen la ecuación son x=5x = 5 y y=2.5y = 2.5.

This problem has been solved

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