Find the solution set of $|z-1| = 1$, where $z\in \mathbb{C}$

Question

Find the solution set of $|z-1| = 1$, where $z\in \mathbb{C}$

This is what I did:

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$$\begin{align} |z-1| = 1 &\implies |z-1|^2 = 1^2 \\ &\implies |(z-1)^2| = 1 \\ &\implies z^2 -2z = 0 \\ &\implies z(z-2) = 0 \\ &\implies x+iy = 0 \ \ \ \ \lor \ \ \ \ x+iy = 2 \ \ \ \\\ &\implies (x = 0 \ \ \land\ \ y = 0) \lor (x=2 \land y = 0) \end{align}$$

Thus the solution to $|z-1| = 1$ are the point $(0,0)$ and $(2, 0)$ on the complex plane.

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EDIT: It is clear that this is not the full solution set? But why does approaching this algebraically as I did, not yield a full solution set? Or has it yielded a full solution set and I've just interpreted things wrong?

Answer 1

Let $z=x+i*y$, where $i^2=-1$. Then we have

$$1=|z-1|=|x+iy-1|=\sqrt{(x-1)^2+y^2},$$

so squaring both sides leads to the equation

$$(x-1)^2+y^2=1.$$

This is the equation of a circle of radius 1, centered at the point $(1,0)$.

Answer 2

$|z-1|=1$ means the distance from the point $(1,0)$ is 1.

Can you write down the equation of circle centered at $(1,0)$ with radius $1$?

Answer 3

You haven't found all solutions.

A different approach is to approach it geometrically. The modulus in your question must be one so this represents a circle with radius of 1 (as you don't care about the argument). The circle is centered where $z-1=0$ hence at $(1,0)$.

Answer 4

Nope, the step of $$|z^2 + 2z + 1| = 1 \implies z^2 + 2z -1 = 0$$ is incorrect. Remember, $|z|$ is the magnitude of $z$.

One way to do this is to write $z = x + iy$, and then solve

$$ |(x + iy) - 1| = 1 \\ |(x - 1) + iy| = 1 \\ (x - 1)^2 + y^2 = 1 \\ (x - 1)^2 + (y - 0)^2 = 1 $$

Which is the equation of a circle of radius $1$ centered at $(1, 0)$

You can imagine this by thinking of a complex number $z$, that is free to sit anywhere in 2D. Then you limit its distance from the point $1 + 0i$ to be $1$, which is exactly the circle of radius 1 centered at $(1, 0)$

Answer 5

Use exponential form: the set $\mathbf U$ of complex numbers with modulus $1$ is simply $\;\bigl\{\mathrm e^{\mathrm i\mkern1.5mu\theta}\mid\theta\in]-\pi,\pi]\bigr\}$. So the solutions are $$\bigl\{1+\mathrm e^{\mathrm i\mkern1.5mu\theta}\mid\theta\in]-\pi,\pi]\bigr\}.$$