In fact, is a domain of and . Then, , with biholomorphic for . Let , then

is -biholomorphic. By a result in [13], and so . Thus, for the domain

the function is -biholomorphic from to a ball of center and of radius equal to . Finally, we remark that implies . Then, if and only if , because We have just seen that there exists a Bloch constant for the class of -holomorphic mappings with on the unit ball and now we wish to estimate this Bloch constant. For this, we need first to work on the natural unit disc: , of bicomplex numbers.

Let and . Then for invertible we obtain: . Also, because and invertible we have: is invertible for near and

Thus, , there exists such that , whenever and is invertible. Choose such that implies . Then,

whenever and both and are invertible. If and invertible but is not, there always exists such that , both and are invertible and Thus, and invertible implies (3.2). This concludes the proof. It is now possible to describe the exact value of the Bloch constant for the class of -holomorphic mappings with on the unit -disc. In fact, it turns out, that it coincides precisely with the classical Bloch constant.

Let us now define:

Then, and so by Theorem 4 and the remark after Theorem 3, with . We want to show that for this , . If not, i.e. , then contains a univalent -disc of radius such that . Thus maps a subdomain biholomorphically onto a -disc of radius . But , so by Lemma 1:

This is a contradiction because Theorem 5 applied to and forces to map the subdomain biholomorphically onto a disc of radius for The followings definitions are used to prove the main result of this article:

Hence inf . Finally, because a -univalent ball for is a univalent ball, we have and then . In fact, by Lemma 1 we have . The second part of the proof is to prove that . We will prove this by contradiction. Suppose that there exists such that . Then

where such that . Hence, there exists, such that and maps a subdomain of biholomorphically onto , a ball centered at of radius . However, by Theorem 8, we know that there exists such that cannot contain a univalent -disc of radius . Also , so . Let us define:

Then . Hence, there exists such that and maps a subdomain of biholomorphically onto . Then, maps the subdomain biholomorphically onto . But , so maps, biholomorphically

This contradicts the way in which g was chosen because with