Find locus of $S$ denoting set of complex numbers $\frac{z+1}{z-3}$, where $z$ varies over set of $|z|=1$.

Question

Question: Let $S$ denote the set of all complex numbers of the form $\frac{z+1}{z-3}$, where $z$ varies over the set of all complex numbers with $|z|=1.$ Find the locus of the points in set $S$.

My approach:

Let $z=x+iy$, with $x,y\in\mathbb{R}$ such that $|z|=1\implies |z|^2=1$. This implies that we must have $x^2+y^2=1$. Now $z+1=(x+1)+iy$ and $z-3=(x-3)+iy$.

Thus \begin{align*} \frac{z+1}{z-3}&=\frac{(z+1)(\overline{z-3})}{|z-3|^2}\\ &=\frac{(z+1)(\overline{z}-3)}{|z-3|^2}\\ &=\frac{x^2-2x-3+y^2-4iy}{(x-3)^2+y^2}\\ &=\frac{-2x-2-4iy}{10-6x}\\ &=\frac{-x-1-2iy}{5-3x}. \end{align*}

Thus we have $$ \Re\left(\frac{z+1}{z-3}\right)=\frac{x+1}{3x-5} $$ and $$ \Im\left(\frac{z+1}{z-3}\right)=\frac{2y}{3x-5},$$ and our task is to find a relationship between these two given that $|z|=1$.

For our ease let us have $\alpha=\frac{z+1}{z-3}\,\, \forall z$ satisfying $|z|=1$.

Now we obtain two useful information using the triangle inequality. We have

$$ |z+1|\le |z|+1=2 $$ and

$$ |z-3|\le |z|+3=4. $$

From here we can conclude that $$|z+1|^2=(x+1)^2+y^2=2+2x\le 4 \iff 1+x\le 2\text{ and } |z-3|^2=5-3x\le 16.$$

Thus we have $$0\le 1+x\le 2 \text{ and } 0\le 5-3x\le 16\implies -1\le x\le 1.$$

Observe that even from $x^2+y^2=1$, we can directly conclude that $-1\le x,y\le 1$.

But this doesn't help much to find a relationship between $\Re(\alpha)$ and $\Im(\alpha)$.

How to proceed?

Answer 1

Let $w=\frac{z+1}{z-3}\implies z=\frac{3w+1}{w-1}$. Then, the given $|z|^2=1$ leads to

$$\frac{3w+1}{w-1}\frac{3\bar w+1}{\bar w-1}=1\implies |w|^2+\frac12(w+\bar w)=0$$

which is just

$$|w+\frac12|^2=\frac1{2^2}$$

Thus, the locus is a circle with center at $-\frac12$ and radius $\frac12$.

Answer 2

Write $w = 1 +\frac{4}{z-3}$, then recognize that this transformation is the composition of 3 transformations: a shift to the left by 3, inversion with a scale factor of 4, then a shift to the right by 1. The tricky part for you is to show that applying inversion to a circle (that does not pass through the origin) yields another circle. Once you do that, you will find that the locus you are looking for is a circle with radius 0.5 centered on $z=-0.5$.

(BTW, this is an example of a Mobius transformation. The neat thing about such transformations is that they preserve the class of circles and lines. Since you start with a circle, you will either get another circle or a line. A line results when the original circle passes through the pole of the transformation. Knowing this, all you need to do is compute a few image points and you can deduce the image locus. For example, $-1\rightarrow 0$, $1\rightarrow -1$, and $i\rightarrow -0.2-0.4i$. You can confirm that $0, -1$ and $-0.2-0.4i$ lie on the aforementioned circle.)

Answer 3

As @Richard said, write $w=1+ \frac{4}{z-3}$

$\Rightarrow z-3=\frac {4}{w-1}$

$\Rightarrow z=\frac {3w+1}{w-1}$

$|z|=|\frac {3w+1}{w-1}|=1$

$|\frac {w+\frac{1}{3}}{w-1}|=\frac{1}{3} \ne1$

Thus, w lies on a circle

Answer 4

You should have proceeded with your approach. You have a calculation mistake which must be fixed first. The real part of $(z+1)/(z-3)$ should be $(x+1)/(3x-5)$ and not $(x+2)/(3x-5)$.

Let $$X=\frac{x+1}{3x-5},Y=\frac{2y}{3x-5}$$ so that $$x=\frac{5X+1}{3X-1},y=\frac{4Y}{3X-1}$$ and we are given $x^2+y^2=1$ so that $$(5X+1)^2+16Y^2=(3X-1)^2$$ ie $$X^2+Y^2+X=0$$ ie $$\left(X+\frac{1}{2}\right)^2+Y^2=\frac{1}{4}$$ which is a circle with center $(-1/2,0)$ and radius $1/2$.

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In general it is better to avoid writing $z=x+iy$ and instead learn a bit of geometry on the complex plane directly. Here is a fact related to circles. If $a, b$ are distinct complex numbers then fixing the argument of $(z-a) /(z-b) $ gives us a circle (a point on a circle subtends the same angle from two given points on the circle). What is not so obvious is that fixing the modulus of $(z-a) /(z-b) $ also gives us a circle (or in special case of modulus $1$ a line the perpendicular bisector of line segment joining $a, b$). This is what is needed here as shown in one of answers.