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

The answer to (1-i)/(1+ki)=2-3i is k= 1/13+8/13 i

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(1-i)/(1+ki)=2-3i

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

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

Solution steps

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

Substitute

k = 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

k = x + y i

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

z - \frac{1}{z} = 3 + 5 i

z^{2} - 3 i z + 5 i = 0

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

x, cs in y + x^{2} y = c

(x + 2 y + 3) + i (3 x - y - 1) = 0

What is the answer to (1-i)/(1+ki)=2-3i ?
The answer to (1-i)/(1+ki)=2-3i is k= 1/13+8/13 i