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

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

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|z+2-i|=2

∣ z + 2 - i ∣ = 2

z = x + (1 + - x (x + 4)) i, z = x + (1 - - x (x + 4)) i; x \in R

Solution steps

∣ z + 2 - i ∣ = 2

Substitute

z = x + y i

∣ x + y i + 2 - i ∣ = 2

Expand

∣ x + y i + 2 - i ∣ : x^{2} + 4 x + y^{2} + 5 - 2 y

x^{2} + 4 x + y^{2} + 5 - 2 y = 2

Solve for

y : y = 1 + - x (x + 4), y = 1 - - x (x + 4)

Substitute back

z = x + y i

z = x + (1 + - x (x + 4)) i, z = x + (1 - - x (x + 4)) i; x \in R

(x + y i) + (8 - 5 i) = 4 + 3 i

z = 9 e^{i \frac{4 \cdot π}{3}}

z^{2} + 8 - i 6 = 0

\frac{5 + i 2}{( - 1 + i 6 ) x} = - 1

5 i (x + i y) + 3 (x - i y) + i y = - 4 - 5 i

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