91Ó°ÊÓ

Let \(f\) and \(g\) be two arbitrary isometries in the complex plane. We will use these isometries to show that their composition is also an isometry. An isometry can be written in the form of \(f(z) = az + b\) where \(a, b \in \mathbb{C}\) and \(|a| = 1\) to ensure that the transformation does not change the distance between points. Remember that for any points \(z_1, z_2 \in \mathbb{C}\), the distance between them is given by \(|z_1 - z_2|\).

Define the composition of the two isometries, say \(h(z)\), by: $$h(z) = f(g(z))$$ Now, let's apply \(f\) and \(g\) to get the composition \(h(z)\): $$h(z) = f(g(z)) = a_1(a_2z + b_2) + b_1$$ Where \(a_1, b_1\) and \(a_2, b_2\) are the complex constants associated with the isometries \(f\) and \(g\), respectively.

To show that \(h(z)\) is an isometry, we need to show that it preserves distances. Consider two points, \(z_1\) and \(z_2\), in the complex plane: $$|h(z_1)-h(z_2)| = |f(g(z_1))-f(g(z_2))|$$ Apply \(f\) to the distances between \(g(z_1)\) and \(g(z_2)\): $$|f(g(z_1))-f(g(z_2))| = |a_1(g(z_1) - g(z_2))|=$$ $$|a_1|\cdot|g(z_1)-g(z_2)|$$ Since \(f\) is an isometry, \(|a_1| = 1\), so $$|a_1|\cdot|g(z_1)-g(z_2)| = |g(z_1)-g(z_2)|$$ Now, apply \(g\) to the distances between \(z_1\) and \(z_2\): $$|g(z_1)-g(z_2)| = |a_2(z_1 - z_2)|$$ Since \(g\) is an isometry, \(|a_2| = 1\), so $$|a_2(z_1 - z_2)| = |z_1 - z_2|$$ Finally, combining the results from applying \(f\) and \(g\), we get: $$|h(z_1)-h(z_2)| = |z_1 - z_2|$$

Since the distance between any two points \(z_1\) and \(z_2\) in the complex plane is preserved under the composition transformation, we can conclude that the composition of two isometries \(f\) and \(g\) is an isometry.