# What are all solutions of |z-1|.|z-1|=1?

*print*Print*list*Cite

### 1 Answer

You need to solve for `z` the equation `|z - 1|^2 = 1` , hence, considering the complex number `z = x + i*y` and replacing it for `z` in equation, yields:

`|x + i*y - 1|^2 = 1`

You need to use the formula of absolute value of a complex number `z = x + i*y` , such that:

`|z| = sqrt(x^2 + y^2) => |z|^2 = x^2 + y^2`

Reasoning by analogy, you may evaluate `|z - 1|^2` , such that:

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

**Hence, evaluating the solutions to the given equation yields that they represent all the points `(x,y)` located on the circle whose center is at `(1,0)` and it has a radius of `1` .**

**Sources:**