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+(1-2sqrt(-x^2+4))/2 i;\quad x\in \mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

|2z-i|=4

∣ 2 z - i ∣ = 4

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

Substitute back

z = x + y i

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

v = i r

z^{2} + 2 z + 1 - i = 0

(5 - 2 i) - 7 = x - (3 + y i)

\frac{1 - i}{1 + ki} = 2 - 3 i

z - \frac{1}{z} = 3 + 5 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+(1-2sqrt(-x^2+4))/2 i;\quad x\in \mathbb{R}