# Find z (complex number) which verify the equality: |z-i|=|z+i*z|=|z-1|

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### 1 Answer

Put the complex number `z=x+i*y`

The module of complex number is `|z| = sqrt(x^2+y^2)`

Calculate `z-i =x + i*y - i`

Factor i: `z-i =x + i*(y-1) =gt |z-i| = sqrt(x^2+(y-1)^2)`

Calculate `z +i*z = z*(1+i) = (1+i)*x + (1+i)*i*y`

`z +i*z = x + i*x + i*y + i^2*y`

Use complex number theory:`i^2 = -1`

`z +i*z = x -y + i*(x+y) =gt |z +i*z| = sqrt((x-y)^2 + (x+y)^2)`

Calculate `z-1 = (x-1) + i*y =gt |z-1| = sqrt((x-1)^2 + y^2)`

Write the condition of problem`:|z-i| = |z +i*z| = |z-1|` =>

`sqrt(x^2+(y-1)^2)=sqrt((x-y)^2 + (x+y)^2)=sqrt((x-1)^2 + y^2)`

Raise to square to remove the square root all over:

`x^2+(y-1)^2=(x-y)^2 + (x+y)^2=(x-1)^2 + y^2`

Expand the squares:

`x^2 + y^2 - 2y + 1 = x^2 - 2xy + y^2 + x^2 + 2xy + y^2 = x^2 - 2x + 1 + y^2`

`x^2 + y^2 - 2y + 1 = 2x^2 + 2y^2 = x^2 + y^2 - 2x + 1`

Equate the first and the last equations:

`x^2 + y^2 - 2y + 1 = x^2 + y^2 - 2x + 1` => 2y = 2x => y=x

Equate the second and the last equations:

`2x^2 + 2y^2 = x^2 + y^2 - 2x + 1`

Subtract the sum `x^2 + y^2` both sides:

`x^2 + y^2 = - 2x + 1`

But x = y => `2x^2 = -2x + 1`

Subtract -2x + 1 both sides:

`2x^2 + 2x- 1 = 0`

Use the quadratic formula:

`x_(1,2) = (-2+-sqrt(4+8))/4 `

`x_(1,2) = (-2+-2sqrt(3))/4`

`x_(1,2) = (-1+-sqrt(3))/2`

Since x=y => `y_(1,2) = (-1+-sqrt(3))/2`

**ANSWER: There are 4 complex numbers z that may be formed from the combinations of `x_(1,2) and y_(1,2).` **

**`z_1 = x_1 + i*y_1` **

**`z_2 = x_1 + i*y_2`**

**`` ****`z_3 = x_2 + i*y_1`**

**`` ****`z_4 = x_2 + i*y_2` **