The following examples will clearly show that a -holomorphic mapping is not necessarily quasiregular.

where In particular, if , then is conformal. However, we note in this example, that is a linear transformation which is nondegenerate if and only if is invertible.

is -holomorphic with . First, we seek conditions on and such that is -quasiregular. By Remark 1 it is easy to show that is -quasiregular on iff

on , that is

Now, , so in this case and . Hence (4.1) becomes

Let . Then, because , we have that . Moreover, and then , i.e. . In Example 2, because , we know by Theorem 6 that . Since the mapping is -quasiregular, we already know that by Theorem 2. However, Theorems 2 and 6 merely assert the existence of the constants and without giving any estimates for these constants. From Theorem 9, on the other hand, we have an interesting lower estimate , by invoking lower estimates on the classical Bloch constant [6]. One can also give lower estimates for the Bloch constant for -quasiregular holomorphic mappings [7]. The next example is a mapping for which by Theorem 6, but for which it is impossible to invoke Theorem 2.

However, , where can be taken to be positive. Finally we see that the last expression goes to infinity as . Finally, we give an example of a mapping which is -holomorphic and biholomorphic without being quasiregular on the unit ball of . In fact, we show that the class of -biholomorphic mappings cannot be totaly included in the class of quasiconformal mappings on the unit ball.