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+(-sqrt(-x^2+1)-2)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 + (- - x^{2} + 1 - 2) 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 = - - x^{2} + 1 - 2

Substitute back

z = x + y i

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

3 z + w = 9 + 11 i

3 - 2 i = \frac{5 x - i}{x + y i}

(2 - i) z + 8 z^{2} = 0

r (t) = (2 t^{2} - 5 t) i + (4 t - 1)

z^{2} = \frac{- 2}{3 i}

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+(-sqrt(-x^2+1)-2)i;\quad x\in \mathbb{R}