Given z=(a+tb)/(c+td) , where t is a real parameter and…

Given $z=\dfrac{a+tb}{c+td}$, where $t$ is a real parameter and $a,b,c,d$ are complex numbers, show that the point $z$ describes a circle for $t\in(-\infty,\infty)$, except in a case that should be specified.

Step 1 · Rewrite $z$ as an affine function of $1/(c+td)$

Assume $d\neq0$. Write:

$$ a+tb = \frac{b}{d}(c+td) + \left(a-\frac{bc}{d}\right) = \frac{b}{d}(c+td) + \frac{ad-bc}{d} $$

Dividing by $c+td$:

$$ z = \frac{a+tb}{c+td} = \frac{b}{d} + \frac{ad-bc}{d}\cdot\frac{1}{c+td} $$

Step 2 · The line traced by $c+td$

As $t$ ranges over $\mathbb{R}$, $\ell(t)=c+td$ traces the straight line $L$ through $c$ with direction $d$ (all of $L$ if $d\neq0$).

Step 3 · Invert the line

The map $w=1/\zeta$ (inversion) sends lines and circles to lines and circles; specifically:

• If $L$ does not pass through the origin (i.e. $c/d\notin\mathbb{R}$), its image $w(t)=1/(c+td)$ is a circle (not through $0$).
• If $L$ does pass through the origin (i.e. $c/d\in\mathbb{R}$, meaning $c$ and $d$ are real multiples of one another), its image is a line through the origin.

Step 4 · Apply the affine map

Since $z = \dfrac{b}{d} + \dfrac{ad-bc}{d}\cdot w$, $z$ is obtained from $w$ by scaling by $\alpha=\dfrac{ad-bc}{d}$ and translating by $\dfrac{b}{d}$. An affine map with $\alpha\neq0$ sends circles to circles and lines to lines (preserving which kind); if $\alpha=0$, every value collapses to the single point $\dfrac{b}{d}$.

Step 5 · Classification

(If $d=0$, $c\neq0$: $z=\dfrac{a+tb}{c}$ is itself affine in $t$, tracing a line if $b\neq0$, or the single point $a/c$ if $b=0$ — consistent with the same classification, since $ad-bc=-bc$ there.)

Step 6 · Why $ad=bc$ is exactly the degenerate case

If $ad=bc$ (and $c\neq0$), then $b=\dfrac{ad}{c}$, so:

$$ z = \frac{a+t\frac{ad}{c}}{c+td} = \frac{a(c+td)}{c(c+td)} = \frac{a}{c} $$

for every $t$ with $c+td\neq0$ — the point never moves.

Step 7 · Algebraic cross-check

Writing $z=x+iy$ and requiring $t=\dfrac{a-zc}{zd-b}$ to be real (i.e. equal to its own conjugate) leads, after clearing denominators, to an equation of the form

$$ \text{Im}(\bar c d)\,(x^2+y^2) + (\text{linear in } x,y) + \text{Im}(\bar a b) = 0, $$

which is a circle when the coefficient $\text{Im}(\bar c d)\neq0$ (equivalently $c/d\notin\mathbb{R}$) and a line when $\text{Im}(\bar c d)=0$ — matching Step 5 exactly, since a circle equation of this form has radius $0$ (a single point) precisely when $ad=bc$.

Step 8 · Numerical check

Take $a=1,\,b=2i,\,c=1,\,d=i$: here $ad-bc = i-2i = -i\neq0$ and $c/d=1/i=-i\notin\mathbb{R}$, so the locus should be a genuine circle. Indeed for $t=0,1,-1$: $z=1,\ \dfrac{1+2i}{1+i}=\dfrac{3+i}{2},\ \dfrac{1-2i}{1-i}=\dfrac{3-i}{2}$ — three distinct, non-collinear points, consistent with a circle (not a line or a point).

Take $a=i,\,b=1,\,c=1,\,d=-i$ (so $ad=i(-i)=1=bc=1\cdot1$): here $z(t)=\dfrac{i+t}{1-it}=i$ for every $t$ — confirming the degenerate, single-point case.