SOLUTION: z is a complex number such that the ratio z-i/z-1 is purely imaginary. Prove that z lies on a circle whose centre is at a point 1/2 + 1/2 i and whose radius is 1/v2( 1 divided by

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: z is a complex number such that the ratio z-i/z-1 is purely imaginary. Prove that z lies on a circle whose centre is at a point 1/2 + 1/2 i and whose radius is 1/v2( 1 divided by      Log On


   

Question 682002: z is a complex number such that the ratio z-i/z-1 is purely imaginary. Prove that z lies on a circle whose centre is at a point 1/2 + 1/2 i and whose radius is 1/v2( 1 divided by square root of 2)
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Since the ratio %28z-i%29%2F%28z-1%29 is purely imaginary,

%28z-i%29%2F%28z-1%29 = ki

z - i = ki(z - 1)

Substitute z = x + yi

x + yi - i = ki(x + yi - 1)

x + yi - i = kxi + kyi² - ki

x + yi - i = kxi + ky(-1) - ki

x + yi - i = kxi - ky - ki

Equate real parts:

(1)     x = -ky

Equate imaginary parts:

yi - i = kxi - ki

y - 1 = kx - k 

(2)    y = kx + 1 - k

Substitute in (1)

      x = -k(kx + 1 - k)

      x = -k²x - k + k²

k²x + x = k² - k

x(k² + 1) = k² - k

        x = %28k%5E2-k%29%2F%28k%5E2%2B1%29

Substitute in (1)

        %28k%5E2-k%29%2F%28k%5E2%2B1%29 = -ky

        %28k%28k-1%29%29%2F%28k%5E2%2B1%29 = -ky

Divide both sides by -k

        %28-%28k-1%29%29%2F%28k%5E2%2B1%29 = y

        %28-k%2B1%29%2F%28k%5E2%2B1%29 = y

We need to show that the point (x,y) = (%28k%5E2-k%29%2F%28k%5E2%2B1%29, %28-k%2B1%29%2F%28k%5E2%2B1%29)

lies on the circle mentioned.    


The point 1%2F2+1%2F2i is the point (1%2F2%29, 1%2F2)

The circle with center (1%2F2%29, 1%2F2) and radius 1%2Fsqrt%282%29

is 

{x - 1%2F2)² + (y - 1%2F2)² = %281%2Fsqrt%282%29%29%5E2

x² - x + 1%2F4 + y² - y + 1%2F4 = 1%2F2

x² - x + y² - y = 0

Substitute (x,y) = (%28k%5E2-k%29%2F%28k%5E2%2B1%29, %28-k%2B1%29%2F%28k%5E2%2B1%29)

%28%28k%5E2-k%29%2F%28k%5E2%2B1%29%29%5E2 - %28k%5E2-k%29%2F%28k%5E2%2B1%29 + %28%28-k%2B1%29%2F%28k%5E2%2B1%29%29%5E2 - %28-k%2B1%29%2F%28k%5E2%2B1%29 = 0

%28k%5E2-k%29%5E2%2F%28k%5E2%2B1%29%5E2 - %28k%5E2-k%29%2F%28k%5E2%2B1%29 + %28-k%2B1%29%5E2%2F%28k%5E2%2B1%29%5E2 - %28-k%2B1%29%2F%28k%5E2%2B1%29 = 0

Get the LCD by multiplying the 2nd and 4th terms by %28k%5E2%2B1%29%2F%28k%5E2%2B1%29

%28k%5E2-k%29%5E2%2F%28k%5E2%2B1%29%5E2 - %28%28k%5E2-k%29%28k%5E2%2B1%29%29%2F%28%28k%5E2%2B1%29%28k%5E2%2B1%29%29 + %28-k%2B1%29%5E2%2F%28k%5E2%2B1%29%5E2 - %28%28-k%2B1%29%28k%5E2%2B1%29%29%2F%28%28k%5E2%2B1%29%28k%5E2%2B1%29%29 = 0

%28k%5E4-2k%5E3%2Bk%5E2%29%2F%28x%5E2%2B1%29%5E2 - %28k%5E4-k%5E3%2Bk%5E2-k%29%2F%28k%2A2%2B1%29%5E2 + %28k%5E2-2k%2B1%29%2F%28k%5E2%2B1%29%5E2 - %28-k%5E3%2Bk%5E2-k%2B1%29%2F%28k%5E2%2B1%29%5E2 = 0

 = 0

0%2F%28k%5E2%2B1%29%5E2 = 0

0 = 0

Therefore the point (x,y) = (%28k%5E2-k%29%2F%28k%5E2%2B1%29, %28-k%2B1%29%2F%28k%5E2%2B1%29)
 lies on the circle with center (1%2F2%29, 1%2F2) and radius 1%2Fsqrt%282%29 

Edwin