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

The answer to (2-i)/(z-1)=2+i is z= 8/5-4/5 i

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

\frac{2 - i}{z - 1} = 2 + i

z = \frac{8}{5} - \frac{4}{5} i

Solution steps

\frac{2 - i}{z - 1} = 2 + i

Substitute

z = x + y i

\frac{2 - i}{x + y i - 1} = 2 + i

Multiply both sides by

x - 1 + i y

\frac{2 - i}{x + y i - 1} (x - 1 + i y) = 2 (x - 1 + i y) + i (x - 1 + i y)

Expand

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

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

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

[2 = 2 x - 2 - y - 1 = - 1 + x + 2 y] : x = \frac{8}{5}, y = - \frac{4}{5}

Substitute back

z = x + y i

z = \frac{8}{5} - \frac{4}{5} i

(x + y) + (x + y) i = 5 + 3 i

z^{2} - 2 i z - 1 = 0

1 - 27 k_{1} i - 8 k_{2} = 0

z^{2} - (2 + i) \cdot z + (- 1 + 7 i) = 0

z, z^{2} = 1 + i

What is the answer to (2-i)/(z-1)=2+i ?
The answer to (2-i)/(z-1)=2+i is z= 8/5-4/5 i