What is the answer to |Z+1|=(Z+2)(1+i) ?

The answer to |Z+1|=(Z+2)(1+i) is Z=(-2+(-1+sqrt(3))/2)-(-1+sqrt(3))/2 i

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|Z+1|=(Z+2)(1+i)

∣ Z + 1 ∣ = (Z + 2) (1 + i)

Z = (- 2 + \frac{- 1 + 3}{2}) - \frac{- 1 + 3}{2} i

Solution steps

∣ Z + 1 ∣ = (Z + 2) (1 + i)

Substitute

Z = x + y i

∣ x + y i + 1 ∣ = (x + y i + 2) (1 + i)

Expand

x^{2} + 2 x + 1 + y^{2} = (x + 2 - y) + i (2 + x + y)

Rewrite

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

in standard complex form:

x^{2} + 2 x + 1 + y^{2} + 0 i

x^{2} + 2 x + 1 + y^{2} + 0 i = (x + 2 - y) + i (2 + x + y)

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[x^{2} + 2 x + 1 + y^{2} = x + 2 - y 0 = 2 + x + y]

[x^{2} + 2 x + 1 + y^{2} = x + 2 - y 0 = 2 + x + y] : (x = - 2 + \frac{- 1 + 3}{2}, y = - \frac{- 1 + 3}{2})

Substitute back

Z = x + y i

Z = (- 2 + \frac{- 1 + 3}{2}) - \frac{- 1 + 3}{2} i

1 + 2 i = z + \frac{i}{z}

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

2 x + 2 y = 2 x + 2 y i + 10, x, y

x - y + i x y = i

z^{2} + (2 + 5 i) \cdot z + (- 6 + 6 i) = 0

What is the answer to |Z+1|=(Z+2)(1+i) ?
The answer to |Z+1|=(Z+2)(1+i) is Z=(-2+(-1+sqrt(3))/2)-(-1+sqrt(3))/2 i