Question: Suppose there is a complex number $\mu$ and a unit vector $u \in V$ (where V is a complex inner product space) such that for all $v \in V$, $\phi(v)=v+(\mu -1)\langle u,v \rangle u.$ Prove that $\phi$ is unitary if and only if $\mu \bar{\mu}=1$.
Solution: $\phi$ is unitary iff $\langle \phi(v),\phi(w) \rangle=\langle v,w \rangle$ $\forall v,w \in V$.
Now,
$$\langle v+(\mu-1)\langle u,v\rangle u,w+(\mu-1) \langle u,w\rangle u\rangle$$
$$=\langle v,w \rangle + (\bar{\mu}-1)\overline{\langle u,v\rangle}\langle u,w\rangle+(\mu-1)\overline{\langle u,v\rangle}\langle u,w\rangle+(\bar{\mu}-1)(\mu-1)\overline{\langle u,v\rangle}\langle u,w\rangle=0$$ so $\mu \bar{\mu}-1=0$.
The part that is puzzling me: This property that was used (although it is not shown):$$\langle v,(\mu -1)\langle u,w\rangle u\rangle =(\bar{\mu}-1)\overline{\langle u,v\rangle}\langle u,w\rangle$$
However, I do not understand this. I understand how the $(\bar{\mu}-1)$ is taken out of the inner product, but I don't understand why $\langle u,w \rangle$ isn't complex conjugate and why $\langle v,u \rangle$ is. I've tried using conjugate symmetry on the whole inner product and then taking the other inner product out, but I still don't get the given result.
Note: $u$ is a unit vector.
Edit: I've posted the whole question and solution to try and make it more clear.
It appears that this computation is done with the assumption that the inner product is conjugate-linear in the first variable, rather than the second variable. So $\langle \lambda u,v\rangle=\bar{\lambda}\langle u,v\rangle$ and $\langle u,\lambda v\rangle=\lambda\langle u,v\rangle$ instead of the other way around. The second term $(\bar{\mu}-1)\overline{\langle u,v\rangle}\langle u,w\rangle$ then comes not from $\langle v,(\mu -1)\langle u,w\rangle u\rangle$ but from $\langle (\mu-1)\langle u,v\rangle u,w\rangle$ (and it is instead the third term that comes from $\langle v,(\mu -1)\langle u,w\rangle u\rangle$).
Indeed, this assumption is necessary for the statement to make sense: notice that $\phi$ isn't even linear if the inner product is conjugate-linear in the second variable. You need to know that if you replace $v$ by $\lambda v$, the term $(\mu-1)\langle u,v\rangle u$ gets multiplied by $\lambda$ (rather than $\bar{\lambda}$).