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

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

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|z-2i|=1

∣ z - 2 i ∣ = 1

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

Solution steps

∣ z - 2 i ∣ = 1

Substitute

z = x + y i

∣ x + y i - 2 i ∣ = 1

Expand

∣ x + y i - 2 i ∣ : x^{2} + y^{2} - 4 y + 4

x^{2} + y^{2} - 4 y + 4 = 1

Solve for

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

Substitute back

z = x + y i

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

x^{2} = - 16 + 30 i

5 x - 5 y i = (x^{2} + y^{2}) (2 + i)

4 x + (y - 3) i = 20

z - \frac{1}{z} = 5 + 3 i

\frac{d}{3.5 h} = 100 \frac{mi}{h}

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