What is the answer to |z-i|=3 ?

The answer to |z-i|=3 is z=x+(1+sqrt(-x^2+9))i,z=x+(1-sqrt(-x^2+9))i;\quad x\in \mathbb{R}

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|z-i|=3

∣ z - i ∣ = 3

z = x + (1 + - x^{2} + 9) i, z = x + (1 - - x^{2} + 9) i; x \in R

Solution steps

∣ z - i ∣ = 3

Substitute

z = x + y i

∣ x + y i - i ∣ = 3

Expand

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

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

Solve for

y : y = 1 + - x^{2} + 9, y = 1 - - x^{2} + 9

Substitute back

z = x + y i

z = x + (1 + - x^{2} + 9) i, z = x + (1 - - x^{2} + 9) i; x \in R

∣ z - i ∣ = 4

(1 + i) x - (1 - 2 i) y = 1 + 4 i

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

x + 2 + 3 y i = - 4 x + (y + 4) i

(2 + i) x - y = 0

What is the answer to |z-i|=3 ?
The answer to |z-i|=3 is z=x+(1+sqrt(-x^2+9))i,z=x+(1-sqrt(-x^2+9))i;\quad x\in \mathbb{R}