This is a homework question that I am struggling with.
I have to normalize the vector $$ \left(\frac{e^{\frac{-i\pi}{4}}}{\sqrt{10}}, -3\frac{e^{\frac{i\pi}{4}}}{\sqrt{10}}\right)$$ as I have to show that it is part of an orthonormal basis. So, I know that the answer should be 1.
When I try to get there, I end up with $$\sqrt{e^{\frac{\pi}{2}}}$$ which obviously, does not equal 1.
I first took the inner product of the vector and, keeping in mind that $ |i| = 1$ I ended up with:
$$\frac{\left(e^{\frac{\pi}{4}}\right)^{2}}{10} + 9 \frac{\left(e^{\frac{\pi}{4}}\right)^{2}}{10} $$
This gave me: $$ 10 \frac{\left(e^{\frac{\pi}{4}}\right)^{2}}{10}$$ which I simplified to: $$\left(e^{\frac{\pi}{4}}\right)^{2}$$ $$=e^{\frac{\pi}{2}}$$
When I take the square root of this to normalize it though, I don't get 1.
I know I've made a mistake, SOMEWHERE, but I can't seem to find it. If anyone could point me in the right direction, it would be much appreciated.
Hint: When taking the inner product between two vector $a,b$ you have to take the complex conjugate of the first, i.e. $$a \cdot b=\overline{a}_ib_i$$
Also remember that $\overline{e^{i\phi}}=e^{-i\phi}$; then you should see that the exponentials cancel.
EDIT (here's how to do it):
Let your vector be $z$. Then $$|z|^2=\overline{z}z=\frac{1}{10}\pmatrix{e^{\frac{i\pi}{4}} \\ -3e^{\frac{-i\pi}{4}}} \pmatrix{e^{\frac{-i\pi}{4}} \\ -3e^{\frac{i\pi}{4}}}=\frac{1}{10}(e^{\frac{i\pi}{4}}e^{\frac{-i\pi}{4}}+9e^{\frac{-i\pi}{4}}e^{\frac{i\pi}{4}})=\frac{1}{10}(1+9)=1$$