What is the answer to |z+2+4i|=2 ?

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

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|z+2+4i|=2

∣ z + 2 + 4 i ∣ = 2

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

Solution steps

∣ z + 2 + 4 i ∣ = 2

Substitute

z = x + y i

∣ x + y i + 2 + 4 i ∣ = 2

Expand

∣ x + y i + 2 + 4 i ∣ : x^{2} + 4 x + y^{2} + 8 y + 20

x^{2} + 4 x + y^{2} + 8 y + 20 = 2

Solve for

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

Substitute back

z = x + y i

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

z \cdot z + 3 (z - z) = 4 - 3 i

x + y i = (3 + i) (2 - 3 i)

z^{2} - 4 i z - 6 = 0

z^{2} - (3 - 2 i) z + (5 - i) = 0

x + y i = (1 - i) (2 + 8 i)

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