What is the answer to |z+i|=3+0i ?

The answer to |z+i|=3+0i is z=x+(-1+sqrt(-x^2+9))i,z=x+(-sqrt(-x^2+9)-1)i;\quad x\in \mathbb{R}

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|z+i|=3+0i

∣ z + i ∣ = 3 + 0 i

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

Solution steps

∣ z + i ∣ = 3 + 0 i

Substitute

z = x + y i

∣ x + y i + i ∣ = 3 + 0 i

Expand

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

Solve for

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

Substitute back

z = x + y i

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

x^{2} - 3 (1 - i) x - 5 i = 0

\frac{( x - 1 )}{2 - x} = i

x y + (x - y) i = 4

∣ z - 1 - i ∣ = 1

x^{2} + 2 (1 + i) x - 5 (1 + 2 i) = 0

What is the answer to |z+i|=3+0i ?
The answer to |z+i|=3+0i is z=x+(-1+sqrt(-x^2+9))i,z=x+(-sqrt(-x^2+9)-1)i;\quad x\in \mathbb{R}