Question

The equation arg `((z - 1)/(z + 1)) = π/4` represents a circle with ______. 

पर्याय

• centre at (0, –1) and radius `sqrt(2)`

• centre at (0, 1) and radius 2

• centre at (0, 1) and radius `sqrt(2)`

• centre at (0, 0) and radius `sqrt(2)`

Answer 1

The equation arg `((z - 1)/(z + 1)) = π/4` represents a circle with `underlinebb(centre  at  (0, 1) and radius sqrt(2))`. 

Explanation:

Let z = x + iy

So, `(z - 1)/(z + 1) = (x + iy - 1)/(x + iy + 1)`

⇒ `(z - 1)/(z + 1) = ((x + iy - 1)/(x + iy + 1)) xx (((x + 1) - iy)/((x + 1) - iy))`

= `((x + 1)(x - 1) - iy(x - 1) + iy(x + 1) - i^2y^2)/((x + 1)^2 + y^2)`

⇒ `(z - 1)/(z + 1) = ((x^2 + y^2 - 1) + i(xy + y - xy + y))/((x + 1)^2 + y^2)`

⇒ `(z - 1)/(z + 1) = ((x^2 + y^2 - 1) + 2iy)/((x + 1)^2 + y^2)`

Given, arg`((z - 1)/(z + 1)) = π/4`

⇒ `(2y)/(x^2 + y^2 - 1) = tan  π/4`  

⇒ `(2y)/(x^2 + y^2 - 1)` = 1

x² + y² – 1 = 2y

x² + y² – 2y – 1 = 0

(x – 0)² + (y – 1)² = `(sqrt(2))^2`

So, Centre (0, 1) and Radius = `sqrt(2)`unit