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

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

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|z-i|=1

∣ z - i ∣ = 1

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

Solution steps

∣ z - i ∣ = 1

Substitute

z = x + y i

∣ x + y i - i ∣ = 1

Expand

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

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

Solve for

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

Substitute back

z = x + y i

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

x^{2} = 4 i

3 x + 2 i y - i x + 5 y = 7 + 5 i

(a + bi)^{2} = - 5 + 12 i

3 + x + i = 7 + y i

- 2 x i + 3 x - 5 y - 10 = - 4 y i - 5 x + 8 i

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