What is the answer to -(i-1)z-w=2+i ?

The answer to -(i-1)z-w=2+i is z=(-1+(w+3)/2)+(w+3)/2 i

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-(i-1)z-w=2+i

- (i - 1) z - w = 2 + i

z = (- 1 + \frac{w + 3}{2}) + \frac{w + 3}{2} i

Solution steps

- (i - 1) z - w = 2 + i

Substitute

z = x + y i

- (i - 1) (x + y i) - w = 2 + i

Expand

- (i - 1) (x + y i) - w : (x + y - w) + i (- x + y)

(x + y - w) + i (- x + y) = 2 + i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[x + y - w = 2 - x + y = 1]

[x + y - w = 2 - x + y = 1] : x = - 1 + \frac{w + 3}{2}, y = \frac{w + 3}{2}

Substitute back

z = x + y i

z = (- 1 + \frac{w + 3}{2}) + \frac{w + 3}{2} i

z^{2} - (5 + i) z + 8 - i = 0

x^{2} + 3 (1 - i) x - 5 i = 0

(2 + 3 i) (x + i y) = (2 + i)^{2}

(9 - 8 i) - 6 = x - (2 + y i)

(x + i y) (4 i) = 8

What is the answer to -(i-1)z-w=2+i ?
The answer to -(i-1)z-w=2+i is z=(-1+(w+3)/2)+(w+3)/2 i