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

The answer to |z+3i|=4 is z=x+(-3+sqrt(-x^2+16))i,z=x+(-sqrt(-x^2+16)-3)i;\quad x\in \mathbb{R}

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|z+3i|=4

∣ z + 3 i ∣ = 4

z = x + (- 3 + - x^{2} + 16) i, z = x + (- - x^{2} + 16 - 3) i; x \in R

Solution steps

∣ z + 3 i ∣ = 4

Substitute

z = x + y i

∣ x + y i + 3 i ∣ = 4

Expand

∣ x + y i + 3 i ∣ : x^{2} + y^{2} + 6 y + 9

x^{2} + y^{2} + 6 y + 9 = 4

Solve for

y : y = - 3 + - x^{2} + 16, y = - - x^{2} + 16 - 3

Substitute back

z = x + y i

z = x + (- 3 + - x^{2} + 16) i, z = x + (- - x^{2} + 16 - 3) i; x \in R

x - 3 i = - 3 + y i

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

∣ (1 + i) z - (1 + 3 i) ∣ = 1

Z = 6 e^{\frac{π}{3} i}

∣ z - 3 i ∣ = 2

What is the answer to |z+3i|=4 ?
The answer to |z+3i|=4 is z=x+(-3+sqrt(-x^2+16))i,z=x+(-sqrt(-x^2+16)-3)i;\quad x\in \mathbb{R}