What is the answer to |z-1+i|=1 ?

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

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|z-1+i|=1

∣ z - 1 + i ∣ = 1

z = (1 + - y (y + 2)) + y i, z = (1 - - y (y + 2)) + y i; y \in R

Solution steps

∣ z - 1 + i ∣ = 1

Substitute

z = x + y i

∣ x + y i - 1 + i ∣ = 1

Expand

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

x^{2} - 2 x + y^{2} + 2 y + 2 = 1

Solve for

x : x = 1 + - y (y + 2), x = 1 - - y (y + 2)

Substitute back

z = x + y i

z = (1 + - y (y + 2)) + y i, z = (1 - - y (y + 2)) + y i; y \in R

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

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

2 y + i y = 3 i + i x - x

x^{2} - 4 y + (2 y - x) i = 2 - i

x^{2} - i = 0

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