Question

solve for z (i-z) (1+2i)+(1-zi)(3-4i)=1+7i​

Answer 1

z = - 1 - i

Step-by-step explanation:

Given,

(i - z) (1 + 2i) + (1 - zi) (3 - 4i) = 1 + 7i

or, {i (1 + 2i) - z (1 + 2i)} + {1 (3 - 4i) - zi (3 - 4i)} = 1 + 7i

or, (i + 2i² - z - 2zi) + (3 - 4i - 3zi + 4zi²) = 1 + 7i

or, (i - 2 - z - 2zi) + (3 - 4i - 3zi - 4z) = 1 + 7i

[ ∵ i² = - 1 ]

or, - 2 - z + i - 2zi + 3 - 4z - 4i - 3zi = 1 + 7i

or, (- 2 - z + 3 - 4z) + (i - 2zi - 4i - 3zi) = 1 + 7i

or, (1 - 5z) + (- 3i - 5zi) = 1 + 7i

or, (1 - 5z) + (- 3 - 5z)i = 1 + 7i

or, 1 - 5 (x + iy) + {- 3 - 5 (x + iy)}i = 1 + 7i

[ ∵ z = x + iy ]

or, 1 - 5x - 5yi - 3i - 5xi - 5yi² = 1 + 7i

or, 1 - 5x - 5yi - 3i - 5xi + 5y = 1 + 7i

or, (1 - 5x + 5y) + (- 5yi - 3i - 5xi) = 1 + 7i

or, (1 - 5x + 5y) + (- 5y - 5x - 3)i = 1 + 7i

Comparing among the real and imaginary parts from both sides, we get

1 - 5x + 5y = 1

or, x = y ..... (1)

and - 5y - 5x - 3 = 7

or, - 5x - 5x - 3 = 7 [ by (1) ]

or, - 10x = 7 + 3

or, 10x = - 10

or, x = - 1

Then y = - 1

Therefore z = - 1 - i