# Batch-Norm makes the Decision Boundary more non-linear?

Consider a Neural Network and let $$L$$ be it's last layer or the output layer. Also suppose we're doing Binary Classification, hence the activation function for the last layer is $$\sigma$$, ie. $$g^{[L]}(.)=\sigma(.)$$. $$\sigma(z^{[L]})=a^{[L]}= \begin{cases} 1, & \text{if}\ a^{[L]}\geq 0.5 \Rightarrow z^{[L]}\geq 0\\ 0, & \text{if}\ a^{[L]}< 0.5 \Rightarrow z^{[L]}< 0 \end{cases}$$

$$z^{[L]}=W^{[L]}a^{[L-1]}+b^{[L]}$$ here $$a^{[L-1]}$$ is a value composed of non-linear activation functions or $$a^{[L-1]}=f(\vec{x}), \vec{x}=[x_1,x_2,...,x_n]$$, the inputs, and hence using the above formulae for $$z^{[L]}$$, $$Z^{[L]}$$ or the Decision Boundary is non-linear! (Note: this is Capital Z which is used to represent for all the training examples, on the contrary Small z is used for only one training example)

Now consider Batch Normalisation, suppose we're using $$\tanh$$ activation function for the hidden layers. Consider two cases, first, the Parameters $$\beta^{[l]}$$ and $$\gamma^{[l]}$$ for most layers are such that most elements of $$Z^{[l]}$$ is lies in the non-linear region, so $$Z^{[L]}$$ or the Decision Boundary will be more non-linear than the second case where $$\beta^{[l]}$$ and $$\gamma^{[l]}$$ for most layers are such that most elements of $$Z^{[l]}$$ lies in the linear part of $$\tanh$$ then the decision boundary will be less non-linear.

Thus Batch-Normalisation makes the Decision Boundary more non-linear, or our model could learn more complex stuff.

Is the above reasoning correct?

Any help is highly appreciated.