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

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

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|z-3i|=3

∣ z - 3 i ∣ = 3

z = x + (3 + - x^{2} + 9) i, z = x + (3 - - x^{2} + 9) i; x \in R

Solution steps

∣ z - 3 i ∣ = 3

Substitute

z = x + y i

∣ x + y i - 3 i ∣ = 3

Expand

∣ x + y i - 3 i ∣ : x^{2} + y^{2} - 6 y + 9

x^{2} + y^{2} - 6 y + 9 = 3

Solve for

y : y = 3 + - x^{2} + 9, y = 3 - - x^{2} + 9

Substitute back

z = x + y i

z = x + (3 + - x^{2} + 9) i, z = x + (3 - - x^{2} + 9) i; x \in R

∣ z - 3 i ∣ = 5

y = in (2 x), y, y = 1, y = 6

x^{2} - 2 x + i = 0

y i n^{0} = 3 y^{2} + x - 1

\frac{s i}{P ( x )} = (x - 1)^{2} + x - 1

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