Let a=α+iβ and z=x+iy correspond to points m(α,β) and M(x,y)…
Let $a=\alpha+i\beta$ and $z=x+iy$ correspond to points $m(\alpha,\beta)$ and $M(x,y)$, related by $z=\dfrac{a-1}{a-2}$.
- Calculate the coordinates of one of the points $m$ or $M$, knowing the coordinates of the other.
- Find the locus of points $M$ when point $m$ describes the line $\alpha=1$.
Explanation of Method
Substitute $a=\alpha+i\beta$ into $z=\dfrac{a-1}{a-2}$ and rationalize by the conjugate of the denominator to get $x,y$ in terms of $\alpha,\beta$; solving the same relation for $a$ in terms of $z$ gives the inverse map. For the locus, set $\alpha=1$, express $x,y$ as functions of the single parameter $\beta$, and eliminate $\beta$ to get an equation in $x,y$ alone.
Part 1a · From $m(\alpha,\beta)$ to $M(x,y)$
Step 1 · Substitute $a=\alpha+i\beta$
$$ z = \frac{(\alpha+i\beta)-1}{(\alpha+i\beta)-2} = \frac{(\alpha-1)+i\beta}{(\alpha-2)+i\beta} $$
Step 2 · Rationalize
Multiply by the conjugate $(\alpha-2)-i\beta$ of the denominator:
$$ z = \frac{[(\alpha-1)+i\beta][(\alpha-2)-i\beta]}{(\alpha-2)^2+\beta^2} $$
Numerator: $(\alpha-1)(\alpha-2)+\beta^2 + i\beta[(\alpha-2)-(\alpha-1)] = (\alpha-1)(\alpha-2)+\beta^2 - i\beta$.
Step 3 · Read off $x$ and $y$
$$ \boxed{x = \frac{(\alpha-1)(\alpha-2)+\beta^2}{(\alpha-2)^2+\beta^2}, \qquad y = \frac{-\beta}{(\alpha-2)^2+\beta^2}} $$
Part 1b · From $M(x,y)$ to $m(\alpha,\beta)$
Step 4 · Solve $z=\dfrac{a-1}{a-2}$ for $a$
$$ z(a-2)=a-1 \quad\Longrightarrow\quad a(z-1) = 2z-1 \quad\Longrightarrow\quad a = \frac{2z-1}{z-1} $$
Step 5 · Substitute $z=x+iy$
$$ a = \frac{(2x-1)+2iy}{(x-1)+iy} $$
Multiply by the conjugate $(x-1)-iy$:
$$ \text{numerator} = [(2x-1)+2iy][(x-1)-iy] = (2x-1)(x-1)+2y^2 + iy\big[2(x-1)-(2x-1)\big] $$
Since $2(x-1)-(2x-1) = 2x-2-2x+1 = -1$, the bracket is simply $-1$, so the numerator is $(2x-1)(x-1)+2y^2 - iy$.
Step 6 · Read off $\alpha$ and $\beta$
$$ \boxed{\alpha = \frac{(2x-1)(x-1)+2y^2}{(x-1)^2+y^2}, \qquad \beta = \frac{-y}{(x-1)^2+y^2}} $$
Part 2 · Locus of $M$ when $\alpha=1$
Step 7 · Substitute $\alpha=1$
With $a=1+i\beta$:
$$ z = \frac{(1+i\beta)-1}{(1+i\beta)-2} = \frac{i\beta}{-1+i\beta} $$
Step 8 · Rationalize
$$ z = \frac{i\beta(-1-i\beta)}{(-1+i\beta)(-1-i\beta)} = \frac{\beta^2-i\beta}{1+\beta^2} $$
Step 9 · Extract $x(\beta)$ and $y(\beta)$
$$ x = \frac{\beta^2}{1+\beta^2}, \qquad y = \frac{-\beta}{1+\beta^2} $$
Step 10 · Eliminate the parameter $\beta$
From $x(1+\beta^2)=\beta^2$: $\ \beta^2 = \dfrac{x}{1-x}$ (for $x\neq1$), so $1+\beta^2=\dfrac{1}{1-x}$.
From $y=\dfrac{-\beta}{1+\beta^2}$: $\ y^2 = \dfrac{\beta^2}{(1+\beta^2)^2}$, so $\dfrac{x}{y^2} = \dfrac{\beta^2/(1+\beta^2)}{\beta^2/(1+\beta^2)^2} = 1+\beta^2 = \dfrac{1}{1-x}$.
Hence $x(1-x)=y^2$, i.e.
$$ x^2+y^2 = x $$
Step 11 · Complete the square
$$ \left(x-\frac12\right)^2+y^2 = \frac14 $$
$$ \boxed{\left(x-\frac12\right)^2+y^2=\frac14} $$
Locus of $M$: a circle of radius $\frac12$ centered at $\left(\frac12,0\right)$, passing through the origin and $(1,0)$.
Step 12 · Verification
Take $\beta=1$ ($\alpha=1$): $x=\frac12$, $y=-\frac12$. Check: $\left(\frac12-\frac12\right)^2+\left(-\frac12\right)^2 = 0+\frac14=\frac14$. ✓
| Quantity | Formula |
|---|---|
| $x(\alpha,\beta)$ | $\dfrac{(\alpha-1)(\alpha-2)+\beta^2}{(\alpha-2)^2+\beta^2}$ |
| $y(\alpha,\beta)$ | $\dfrac{-\beta}{(\alpha-2)^2+\beta^2}$ |
| $\alpha(x,y)$ | $\dfrac{(2x-1)(x-1)+2y^2}{(x-1)^2+y^2}$ |
| $\beta(x,y)$ | $\dfrac{-y}{(x-1)^2+y^2}$ |
| Locus for $\alpha=1$ | $\left(x-\frac12\right)^2+y^2=\frac14$ |
Result
As $m$ moves along the vertical line $\alpha=1$, its image $M$ under $z=\dfrac{a-1}{a-2}$ traces a circle of radius $\frac12$ centered at $\left(\frac12,0\right)$ — a hallmark behavior of Möbius (fractional linear) transformations, which map lines and circles to lines or circles.