Answer to Question #118430 in Complex Analysis for Nii Laryea

Question

The transformation T : z "→" w in the complex plane is defined by w = (az + b)/( z + c), where a, b, c ∈ R. Given that w = 3i when z = −3i, and w = 10 − 4i when z = 1 + 4i, find the values of a, b and c. (a) Show that the points for which z is transformed to z lie on a circle and give the centre and radius of this circle. (b) Show that the line through the point z = 4 and perpendicular to the real axis is invariant under T.

Answer 1

After substitution of $w = 3i,\, z=-3i$ we receive:

$3i=\frac{-3ai+b}{c-3i}\Longrightarrow 3c\,i+9=b-3ai$ .

Since $a,\,b,\,c$ are real, we get $b=9,\,a+c=0$ .

Setting $w=10-4i,\,z=1+4i$ we get:

$10-4i=\frac{(1+4i)a+9}{1+4i-a}\Longrightarrow 26+36i-a(10-4i)=a(1+4i)+9$

The latter implies that $36\,i=0$ .

Thus, there are no coefficients $a,b,c$ satisfying the problem.

(a) We set $w=z$ and receive:

$z=\frac{az+b}{z+c}\Longrightarrow z^2+(c-a)z-b=0,\,\,z\neq-c$

Setting $z=x+i\,\,y$ we get

$x^2+2xy\,i-y^2+(c-a)(x+i\,\,y)-b=0$ .

The latter leads to

$\left\{\begin{array}{l} x^2-y^2+(c-a)x-b=0 \\ 2xy+(c-a)y=0 \end{array}\right.$

Solving the latter, we get

1. If $D=(a-c)^2+4b>0$ then $x=\frac{a-c\pm\sqrt{(a-c)^2+4b}}{2},\quad y=0$ ;
2. If $D=(a-c)^2+4b\leq 0$ then $x=\frac12(a-c),\quad y=\pm\frac12\sqrt{-(a-c)^2-4b}.$

Now we shall construct a circle for each case.

We set $(x-x_0)^2+(y-y_0)^2=R^2$ ,

1. $D=(a-c)^2+4b>0$ . Substitution of points yields the following system

$\left\{\begin{array}{l} (x_1-x_0)^2+y_0^2=R^2 \\ (x_2-x_0)^2+y_0^2=R^2 \end{array}\right.$ with $x_1=\frac{a-c-\sqrt{(a-c)^2+4b}}{2},\,\, x_2=\frac{a-c+\sqrt{(a-c)^2+4b}}{2}$

We subtract the first equation from the second and then solve the obtained equation with respect to $x_0$ . We receive $x_0=\frac{x_1+x_2}{2}=a-c.$ We can then choose an arbitrary $y_0$ and $R=\sqrt{\frac14(x_1-x_2)^2+y_0^2}$ .

2. $D=(a-c)^2+4b\leq 0$ . Substitution of points yields the system:

$\left\{\begin{array}{l} (x_1-x_0)^2+(y_1-y_0)^2=R^2 \\ (x_1-x_0)^2+(-y_1-y_0)^2=R^2 \end{array}\right.$ with $y_1=\frac12\sqrt{-(a-c)^2-4b}.$

After substracting of the first equation from the second we receive that $y_1y_0=0$ .

If $(a-c)^2+4b=0$ we can choose arbitrary $x_0,y_0$ and set $R=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$ . Otherwise, we get $y_0=0$ . We choose an arbitrary $x_0$ and set $R=\sqrt{(x_1-x_0)^2+y_1^2}$ .

(b) All points on the line can be presented as $z=4+y\,i$ , $y\in\mathbb{R}$. After acting with the transformation we receive

$w=\frac{a(4+y\,i)+b}{4+y\,i+c}$ .

We substitute $y=0,\,\,y=1$ and receive $w=\frac{4a+b}{4+c}$ , $w=\frac{(4a+b)(c+4)+1}{(c+4)^2+1}+\frac{c+4-(4a+b)}{(c+4)^2+1}\,i$

In general case (arbitrary $a,b,c$ ), these two points are not of the form $w=4+\tilde{y}\,i$ , ${\tilde{y}}\in{\mathbb{R}}$ .