Find an ortonormal base for the subespace $S$ of $\mathbb{C}^3$ span by the vectors $v_1=(1,0,i)$ and $v_2=(2,1,1+i)$ with the usual inner product and find also a base for the space $S^{\perp}$.
My atemptt
If we compute $<v_1,v_2>$, we obtain
$<v_1,v_2>=<(1,0,i),(2,1,1+i)>=2+i\overline{(1+i)}=2+i(1-i)=2+i-i^2=2+i+1=3+i$.
Then $S$ is not ortonormal yet so we need to use the Gram Schmidt process:
Let $u_1=v_1=(1,0,i)$ and $u_2=v_2-proj_{u_1}(v_2)=(2,1,1+i)-\frac{<(1,0,i),(2,1,1+i)>}{<(1,0,i),(1,0,i)>}(1,0,i)=(2,1,1+i)-\frac{3+i}{2}(1,0,i)=(2,1,1+i)-(\frac{3+i}{2},0,\frac{3i-1}{2})=(2-\frac{3+i}{2},1,1+i-(\frac{3i-1}{2}))=(\frac{1-i}{2},1,\frac{3-i}{2})$.
But after doing this $<u_1,u_2>$ I'm obtaining $<(1,0,i),(\frac{1-i}{2},1,\frac{3-i}{2})>=\frac{1+i}{2}+i\frac{3+i}{2}=\frac{1+i+3i-1}{2}=2i$.
I don't know why, where am I doing wrong?
$proj_{u_1}(v_2)=\frac{\langle v_2,u_1\rangle}{\langle u_1,u_1\rangle}u_1=\frac{\color{red}{\bar{\langle u_1,v_2\rangle}}}{\langle u_1,u_1\rangle}u_1$, with $\langle v_2,u_1\rangle=3\color{red}-i$. You will obtain : $u_2=(\frac{1+i}{2},1,\frac{1-i}{2})$ and finally $\langle u_1,u_2\rangle=0$