What is the answer to -1/(z-i)+1-z=0 ?

The answer to -1/(z-i)+1-z=0 is z=0.25706…+1.52908…i,z=0.74293…-0.52908…i

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-1/(z-i)+1-z=0

- \frac{1}{z - i} + 1 - z = 0

z = 0.25706 \dots + 1.52908 \dots i, z = 0.74293 \dots - 0.52908 \dots i

Solution steps

- \frac{1}{z - i} + 1 - z = 0

Substitute

z = x + y i

- \frac{1}{x + y i - i} + 1 - (x + y i) = 0

Multiply both sides by

x + i (- 1 + y)

- \frac{1}{x + y i - i} (x + i (- 1 + y)) + 1 \cdot (x + i (- 1 + y)) - (x + y i) (x + i (- 1 + y)) = 0 \cdot (x + i (- 1 + y))

Expand

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

[- 1 + x - x^{2} - y + y^{2} = 0 - 1 + x + y - 2 x y = 0]

[- 1 + x - x^{2} - y + y^{2} = 0 - 1 + x + y - 2 x y = 0] : (x = 0.25706 \dots, x = 0.74293 \dots, y = 1.52908 \dots y = - 0.52908 \dots)

Substitute back

z = x + y i

z = 0.25706 \dots + 1.52908 \dots i, z = 0.74293 \dots - 0.52908 \dots i

+ 5 i z^{2} = - 2 + 3 i, z^{1} = 4

z^{1} + z^{2} = (4 + 3) + (2 + 5) i

z^{2} - (1 + 4 i) z - (6 - 12 i) = 0

(2 + i) \cdot a + (i - 3) \cdot b = 4

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

What is the answer to -1/(z-i)+1-z=0 ?
The answer to -1/(z-i)+1-z=0 is z=0.25706…+1.52908…i,z=0.74293…-0.52908…i