What is the answer to |z+i|=2 ?

The answer to |z+i|=2 is z=x+(-1+sqrt(-x^2+4))i,z=x+(-sqrt(-x^2+4)-1)i;\quad x\in \mathbb{R}

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|z+i|=2

∣ z + i ∣ = 2

z = x + (- 1 + - x^{2} + 4) i, z = x + (- - x^{2} + 4 - 1) i; x \in R

Solution steps

∣ z + i ∣ = 2

Substitute

z = x + y i

∣ x + y i + i ∣ = 2

Expand

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

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

Solve for

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

Substitute back

z = x + y i

z = x + (- 1 + - x^{2} + 4) i, z = x + (- - x^{2} + 4 - 1) i; x \in R

4 (4 i + 6) = (4 x) i + 8 y

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

2 a + 2 bi = 2 (a + bi) + 9

\frac{z + 1}{z - 1} = 2 + 3 i

i z + ∣ z ∣ + 3 i = 1

What is the answer to |z+i|=2 ?
The answer to |z+i|=2 is z=x+(-1+sqrt(-x^2+4))i,z=x+(-sqrt(-x^2+4)-1)i;\quad x\in \mathbb{R}