What is the answer to |z+3-4i|=2 ?

The answer to |z+3-4i|=2 is z=x+(4+sqrt(-x^2-6x-5))i,z=x+(4-sqrt(-x^2-6x-5))i;\quad x\in \mathbb{R}

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|z+3-4i|=2

∣ z + 3 - 4 i ∣ = 2

z = x + (4 + - x^{2} - 6 x - 5) i, z = x + (4 - - x^{2} - 6 x - 5) i; x \in R

Solution steps

∣ z + 3 - 4 i ∣ = 2

Substitute

z = x + y i

∣ x + y i + 3 - 4 i ∣ = 2

Expand

∣ x + y i + 3 - 4 i ∣ : x^{2} + 6 x + y^{2} + 25 - 8 y

x^{2} + 6 x + y^{2} + 25 - 8 y = 2

Solve for

y : y = 4 + - x^{2} - 6 x - 5, y = 4 - - x^{2} - 6 x - 5

Substitute back

z = x + y i

z = x + (4 + - x^{2} - 6 x - 5) i, z = x + (4 - - x^{2} - 6 x - 5) i; x \in R

2 x + 9 y i = 16 - 12 i

z^{2} = 96 + 40 i

a - ib = - 1 b + ia

z^{2} - 2 i z - 1 - 3 i = 0

∣ z - 2 i ∣ = 1

What is the answer to |z+3-4i|=2 ?
The answer to |z+3-4i|=2 is z=x+(4+sqrt(-x^2-6x-5))i,z=x+(4-sqrt(-x^2-6x-5))i;\quad x\in \mathbb{R}