How to algebraically determine locus of points of $|z|=|z-2|$?

Question

How would I algebraically determine the locus of points in the $z$-plane that satisfy the equation $|z|=|z-2|$?

Answer 1

put $z=a+\iota b$

$$|z|=|z-2|$$ $$|a+\iota b|=|a+\iota b-2|$$ $$|a+\iota b|=|a-2+\iota b|$$ $$a^2+b^2=(a-2)^2+b^2$$ $$a^2+b^2=a^2+4-4a+b^2$$ $$4=4a$$ gives us $$a=1$$ means equation of the locus is $$Re(z)=1$$

Answer 2

The equation says it,

$$\sqrt{x^2+y^2}=\sqrt{(x-2)^2+y^2}.$$

Squaring won't introduce new solutions as both members are positive and the arguments of the square roots are positive, hence after simplification

$$0=4-4x.$$

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From the equation,

$$z=(z-2)e^{i\theta}$$ and

$$z=-\frac{2e^{i\theta}}{1-e^{i\theta}}=-\frac{2e^{i\theta}(1-e^{-i\theta})}{(1-e^{i\theta})(1-e^{-i\theta})}=\frac{2-2e^{i\theta}}{2-2\Re(e^{i\theta})}$$ which has the real part $1$.

Answer 3

Hint: $\,|z|^2 = |z-2|^2 \iff z \bar z = (z-2)(\bar z - 2) \iff 2z + 2\bar z = 4 \iff \operatorname{Re}(z)=1\,$.

Of course, it's more direct to note that geometrically the locus is the pependicular bisector of the segment between $0$ and $2$.