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

The answer to (1-i)/(1+z*i)=2-3i,z is z= 1/13+8/13 i

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(1-i)/(1+z*i)=2-3i,z

\frac{1 - i}{1 + z \cdot i} = 2 - 3 i, z

z = \frac{1}{13} + \frac{8}{13} i

Solution steps

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

Substitute

z = x + y i

\frac{1 - i}{1 + ( x + y i ) i} = 2 - 3 i

Multiply both sides by

1 + (x + y i) i

\frac{1 - i}{1 + ( x + y i ) i} (1 + (x + y i) i) = 2 (1 + (x + y i) i) - 3 i (1 + (x + y i) i)

Expand

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

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

[1 = 2 - 2 y + 3 x - 1 = - 3 + 2 x + 3 y]

[1 = 2 - 2 y + 3 x - 1 = - 3 + 2 x + 3 y] : x = \frac{1}{13}, y = \frac{8}{13}

Substitute back

z = x + y i

z = \frac{1}{13} + \frac{8}{13} i

z^{2} - (3 - 4 i) \cdot z + (2 - 8 i) = 0

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

∣ z - 3 + 4 i ∣ = 5

(3 - 2 i) (x + i) = 17 + i y

z^{2} + z \cdot i + 2 = 0

What is the answer to (1-i)/(1+z*i)=2-3i,z ?
The answer to (1-i)/(1+z*i)=2-3i,z is z= 1/13+8/13 i