I have the following problem: by having the orthonormal basis vectors $|u_1\rangle, |u_2\rangle$ I need to determine the orthonormal vector of:
$|\psi\rangle = \frac{1}{\sqrt2} (|u_1\rangle + i\cdot|u_2\rangle$
The answer is denoted by:
$((\langle u_1|-i\cdot\langle u_2|)\cdot(\alpha|u_1\rangle+\beta\cdot|u_2\rangle) = 0 \Rightarrow \alpha = i\beta$
$\alpha*\alpha + \beta*\beta = 1 \Rightarrow \beta*\beta = \frac{1}{2} \Rightarrow \beta = \frac{1}{\sqrt2}e^{î\varphi}$
$|\psi^{\perp}\rangle = \frac{1}{\sqrt{2}}(i|u_1\rangle+|u_2\rangle)e^{î\varphi}$
I do not have any idea where does the $\alpha$ or $\beta$ are defined or from which theorem are taken, normally I would take the complex conjugate of $|\psi\rangle$ and do the following:
$|\psi^{\perp}\rangle = \frac{|\psi\rangle}{\sqrt{\langle\psi|\psi\rangle}}$
Which gives a totally different solution as the one proposed, I will be very grateful if somebody could explain this to me. Thank you in advance for your time and consideration!
You are looking for a vector $|\psi^\perp\rangle$ such that $\langle \psi|\psi^\perp\rangle = 0$ (orthogonality to $|\psi\rangle$) and $\langle \psi^\perp |\psi^\perp\rangle=1$ (being normalized). From $|\psi\rangle = \frac 1{\sqrt 2}(|u_1\rangle + i\,|u_2\rangle)$ you have $$\langle\psi| = \frac 1{\sqrt 2}(\langle u_1| - i\,\langle u_2|).$$ Since $|u_1\rangle$, $|u_2\rangle$ form a basis, the vector $|\psi^\perp\rangle$ can be expressed as $|\psi^\perp\rangle = \alpha\,|u_1\rangle+\beta\,|u_2\rangle$ for some scalars $\alpha$ and $\beta$.
Now the two conditions $\langle \psi|\psi^\perp\rangle = 0$ and $\langle \psi^\perp |\psi^\perp\rangle=1$ yield \begin{align*} 0 &= \langle \psi|\psi^\perp\rangle = \frac 1{\sqrt 2}(\langle u_1| - i\,\langle u_2|)(\alpha\,|u_1\rangle+\beta\,|u_2\rangle) = \frac 1{\sqrt 2}(\alpha-i\beta), \\ 1 &= \langle \psi^\perp|\psi^\perp\rangle = (\overline \alpha\,\langle u_1|+\overline \beta\,\langle u_2|)(\alpha\,|u_1\rangle+\beta\,|u_2\rangle) = |\alpha|^2 + |\beta|^2. \end{align*} Hence, $\alpha=i\beta$ and putting that into the second equations yields $2|\beta|^2=1$ so $|\beta|=\frac 1{\sqrt 2}$ which means that $\beta=\frac 1{\sqrt 2}e^{i\varphi}$ for some angle $\varphi$. Plugging back into $\alpha=i\beta$ yields $\alpha=i\frac 1{\sqrt 2}e^{i\varphi}$ and hence $$ |\psi^\perp\rangle = \alpha\,|u_1\rangle+\beta\,|u_2\rangle = \frac 1{\sqrt 2} e^{i\varphi} \left(i\,|u_1\rangle+|u_2\rangle\right). $$