Formula for inner product 
Prove the following formula for inner product $$4 \langle u,v \rangle = ||u+v||^2+i||u+iv||^2-||u-v||^2-i||u-iv||^2$$

Attempted solution:
Note that $$\begin{align}&||u+v||^2+i||u+iv||^2-||u-v||^2-i||u-iv||^2 \\&= \langle u+v, u+v \rangle + i\langle u+iv, u+iv \rangle - \langle u-v, u-v \rangle  - i \langle u-iv, u-iv \rangle \\&= \langle u,u\rangle+\langle u,v\rangle+\langle v,u \rangle+\langle v,v\rangle + i\langle u,u \rangle + i\langle u,iv \rangle + i\langle iv,u \rangle + i\langle iv,iv\rangle - \langle u,u \rangle -\langle u,-v\rangle - \langle -v,u \rangle - \langle -v,-v\rangle - i\langle u,u \rangle - i\langle u,-iv\rangle - i\langle -iv,u\rangle - i\langle -iv,-iv\rangle \\&= 2\langle u,v \rangle  +i \langle v,v \rangle + \langle v,v \rangle \\&\neq4\langle u,v\rangle\end{align}$$
Where do I go wrong?
 A: You assumed that $\langle u,v\rangle=\langle v,u\rangle$; it should be $\langle u,v\rangle=\overline{\langle v,u\rangle}$. Besides, $\langle u,iv\rangle=-i\langle u,v\rangle$. You are working with a sesquilinear form here.
A: Since we are working with a complex inner product we have to use conjugate symmetry: $\langle u,v\rangle=\overline{\langle v,u\rangle}$. This also tells us that $\langle u,cv\rangle=\overline{c}\langle u,v\rangle$. I highlight the step in your calculation where you should have used this conjugate symmetry. Terms with the same colour cancel out.
$$\begin{align}&\color{red}{\langle u,u\rangle}+\langle u,v\rangle+\langle v,u \rangle+\color{blue}{\langle v,v\rangle} + \color{green}{i\langle u,u \rangle} + i\langle u,iv \rangle + i\langle iv,u \rangle + \color{purple}{i\langle iv,iv\rangle} - \color{red}{\langle u,u \rangle} -\langle u,-v\rangle - \langle -v,u \rangle - \color{blue}{\langle -v,-v\rangle} - \color{green}{i\langle u,u \rangle} - i\langle u,-iv\rangle - i\langle -iv,u\rangle - \color{purple}{i\langle -iv,-iv\rangle} \\&=\langle u,v\rangle+\color{blue}{\overline{\langle u,v\rangle}}+i\overline i\langle u,v\rangle+\color{blue}{i^2\overline{\langle u,v\rangle}}+\langle u,v\rangle+\color{red}{\overline{\langle u,v\rangle}}+i\overline i\langle u,v\rangle+\color{red}{i^2\overline{\langle u,v\rangle}}\\&=4\langle u,v\rangle\end{align}$$
Note, we used $i^2=-1$ and $i\overline i=1$.
