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

The answer to |z-(1-i)|=2 is z=(1+sqrt(-y^2-2y+3))+yi,z=(1-sqrt(-y^2-2y+3))+yi;\quad y\in \mathbb{R}

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|z-(1-i)|=2

∣ z - (1 - i) ∣ = 2

z = (1 + - y^{2} - 2 y + 3) + y i, z = (1 - - y^{2} - 2 y + 3) + y i; y \in R

Solution steps

∣ z - (1 - i) ∣ = 2

Substitute

z = x + y i

∣ x + y i - (1 - i) ∣ = 2

Expand

∣ x + y i - (1 - i) ∣ : x^{2} - 2 x + y^{2} + 2 y + 2

x^{2} - 2 x + y^{2} + 2 y + 2 = 2

Solve for

x : x = 1 + - y^{2} - 2 y + 3, x = 1 - - y^{2} - 2 y + 3

Substitute back

z = x + y i

z = (1 + - y^{2} - 2 y + 3) + y i, z = (1 - - y^{2} - 2 y + 3) + y i; y \in R

a - bi = i (a + bi)

∣ z - 1 + i ∣ = 3

\frac{z + 1}{z - 1} = 2 + i^{25}

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

\frac{26 + 28 i}{2 + i} = a + bi

What is the answer to |z-(1-i)|=2 ?
The answer to |z-(1-i)|=2 is z=(1+sqrt(-y^2-2y+3))+yi,z=(1-sqrt(-y^2-2y+3))+yi;\quad y\in \mathbb{R}