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

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

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|2z+i|=4

∣ 2 z + i ∣ = 4

z = x + \frac{- 1 + 2 - x ^{2} + 4}{2} i, z = x - \frac{2 - x ^{2} + 4 + 1}{2} i; x \in R

Solution steps

∣ 2 z + i ∣ = 4

Substitute

z = x + y i

∣ 2 (x + y i) + i ∣ = 4

Expand

∣ 2 (x + y i) + i ∣ : 4 x^{2} + 4 y^{2} + 4 y + 1

4 x^{2} + 4 y^{2} + 4 y + 1 = 4

Solve for

y : y = \frac{- 1 + 2 - x ^{2} + 4}{2}, y = - \frac{2 - x ^{2} + 4 + 1}{2}

Substitute back

z = x + y i

z = x + \frac{- 1 + 2 - x ^{2} + 4}{2} i, z = x - \frac{2 - x ^{2} + 4 + 1}{2} i; x \in R

r, i = \frac{v}{r}

5 x + i 5 y = 100, 5 y + i (5 x - 10 y) = 60 + i 80

x^{2} + (4 - 2 i) x + 3 = 0

y, (2 x - y) - 4 i = 12 + 2

(3 + 3 i) (a + bi) = 3 + 18 i

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