What is the answer to |z-1+3i|=2 ?

The answer to |z-1+3i|=2 is z=(1+sqrt(-y^2-6y-5))+yi,z=(1-sqrt(-y^2-6y-5))+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+3i|=2

∣ z - 1 + 3 i ∣ = 2

z = (1 + - y^{2} - 6 y - 5) + y i, z = (1 - - y^{2} - 6 y - 5) + y i; y \in R

Solution steps

∣ z - 1 + 3 i ∣ = 2

Substitute

z = x + y i

∣ x + y i - 1 + 3 i ∣ = 2

Expand

∣ x + y i - 1 + 3 i ∣ : x^{2} - 2 x + y^{2} + 6 y + 10

x^{2} - 2 x + y^{2} + 6 y + 10 = 2

Solve for

x : x = 1 + - y^{2} - 6 y - 5, x = 1 - - y^{2} - 6 y - 5

Substitute back

z = x + y i

z = (1 + - y^{2} - 6 y - 5) + y i, z = (1 - - y^{2} - 6 y - 5) + y i; y \in R

∣ x + bi + 1 + i ∣ = 1

(z^{2} + i) = 0

4 - x i = y - 3 i

i (t) = t^{2} - 2 t + 1

\frac{Z}{1 + Z} = 3 + i 4

What is the answer to |z-1+3i|=2 ?
The answer to |z-1+3i|=2 is z=(1+sqrt(-y^2-6y-5))+yi,z=(1-sqrt(-y^2-6y-5))+yi;\quad y\in \mathbb{R}