Given $\langle v|v\rangle=\sum_{i}a_{i}^2$ then $|||v\rangle||=\sqrt{\langle v|v\rangle}=\sqrt{\sum_{i}|a_{i}|^2}$.
Also, $\langle a|b\rangle=\langle b|a\rangle^{*}$.
So how then is $\langle a|b\rangle\langle b|a\rangle=||\langle a|b\rangle||^{2}=(\sqrt{\sum_{i}a_{i}^{*}b_{i}})^{2}$?
I think I have misunderstood what the norm is.
The only way I can see the above working is if $||\langle a|b\rangle||^{2}=(\sqrt{\sum_{i}a_{i}^{*}b_{i}})^{2}=\sqrt{\langle a|b\rangle\langle b|a\rangle}$
If this is the case, then I believe I have gotten confused regarding what the norm is actually doing, as I normally only come across it for either complex numbers or just single vectors. If the above does hold, then I take it to mean I need to take the sum of the product of the coefficients and their conjugate for vectors and complex numbers, but for inner products I must take the product of the innerproduct and it's transpose?
Given the inner product $$ \langle a | b \rangle = \sum_ia^*_ib_i $$ we have $$ \langle b | a \rangle = \langle a | b \rangle^* = \sum_ia_ib^*_i $$ then \begin{align} |\langle a | b \rangle|^2 &=\langle b | a \rangle \langle a | b \rangle = \langle a | b \rangle^*\langle a | b \rangle = \\ &= \sum_i a_i b^*_i \sum_j a^*_j b_j = \left( \sum_i a^*_i b_i \right)^* \sum_j a^*_j b_j = \left|\sum_j a^*_j b_j\right|^2 \end{align}