What is the answer to |z-1+i|=3 ?

The answer to |z-1+i|=3 is z=(1+sqrt(-y^2-2y+8))+yi,z=(1-sqrt(-y^2-2y+8))+yi;\quad y\in \mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

|z-1+i|=3

∣ z - 1 + i ∣ = 3

z = (1 + - y^{2} - 2 y + 8) + y i, z = (1 - - y^{2} - 2 y + 8) + y i; y \in R

Solution steps

∣ z - 1 + i ∣ = 3

Substitute

z = x + y i

∣ x + y i - 1 + i ∣ = 3

Expand

∣ x + y i - 1 + i ∣ : x^{2} - 2 x + y^{2} + 2 y + 2

x^{2} - 2 x + y^{2} + 2 y + 2 = 3

Solve for

x : x = 1 + - y^{2} - 2 y + 8, x = 1 - - y^{2} - 2 y + 8

Substitute back

z = x + y i

z = (1 + - y^{2} - 2 y + 8) + y i, z = (1 - - y^{2} - 2 y + 8) + y i; y \in R

\frac{z + 1}{z - 1} = 2 + i^{25}

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

\frac{26 + 28 i}{2 + i} = a + bi

a + bi = 9 + 8 i

2 i x + 3 = y - i

What is the answer to |z-1+i|=3 ?
The answer to |z-1+i|=3 is z=(1+sqrt(-y^2-2y+8))+yi,z=(1-sqrt(-y^2-2y+8))+yi;\quad y\in \mathbb{R}