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

The answer to (1-i)z+w=3+2i is z=(-2+(-w+5)/2)+(-w+5)/2 i

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(1-i)z+w=3+2i

(1 - i) z + w = 3 + 2 i

z = (- 2 + \frac{- w + 5}{2}) + \frac{- w + 5}{2} i

Solution steps

(1 - i) z + w = 3 + 2 i

Substitute

z = x + y i

(1 - i) (x + y i) + w = 3 + 2 i

Expand

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

(x + y + w) + i (- x + y) = 3 + 2 i

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

[x + y + w = 3 - x + y = 2]

[x + y + w = 3 - x + y = 2] : x = - 2 + \frac{- w + 5}{2}, y = \frac{- w + 5}{2}

Substitute back

z = x + y i

z = (- 2 + \frac{- w + 5}{2}) + \frac{- w + 5}{2} i

z^{2} - (3 - 4 i) z - (1 + 7 i) = 0

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

i = \frac{e}{r}

(3 x - 4 y) + (6 x + 2 y) i = 5 i

z^{2} = 1 - 3 i

What is the answer to (1-i)z+w=3+2i ?
The answer to (1-i)z+w=3+2i is z=(-2+(-w+5)/2)+(-w+5)/2 i