Verify that $$v_1 = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right), v_2=\left(\frac 1{\sqrt{3}},\frac{1}{\sqrt{3}}e^{-2i\pi/3},\frac{1}{\sqrt{3}}e^{2i\pi/3}\right), v_3=\left(\frac1{\sqrt{3}},\frac{1}{\sqrt{3}}e^{2i\pi/3},\frac{1}{\sqrt{3}}e^{-2i\pi/3}\right)$$ is an orthonormal basis for the complex inner product space $\Bbb C^3$.
I used the dot product as the standard inner product and assumed that the complex numbers were written in euler notation. I started by looking at the norms of each vector to make sure that they were 1. The norm of the first vector was 1, but when I looked at the second vector I had his problem:
$$\left(\frac 1{\sqrt{3}}, -\frac 1{2\sqrt{3}}-\frac i2,-\frac 1{2\sqrt{3}}+\frac i2\right)\cdot \left(\frac 1{\sqrt{3}}, -\frac 1{2\sqrt{3}}-\frac i2,-\frac 1{2\sqrt{3}}+\frac i2\right) = 0$$
so when I tried to find the square of the norm I got zero. Can anybody help me see where I have gone wrong? I have triple checked everything and still keep getting zero. I have the same problem with the third vector as the second two terms are just swapped.
$$\begin{align}\|v_2\|^2 &= v_2\cdot v_2 \\ &=\left(\frac 1{\sqrt{3}},\frac{1}{\sqrt{3}}e^{-2i\pi/3},\frac{1}{\sqrt{3}}e^{2i\pi/3}\right)\cdot \left(\frac 1{\sqrt{3}},\frac{1}{\sqrt{3}}e^{-2i\pi/3},\frac{1}{\sqrt{3}}e^{2i\pi/3}\right) \\ &= \left(\overline{\frac 1{\sqrt{3}}}\cdot \frac 1{\sqrt{3}}\right) + \left(\overline{\frac{1}{\sqrt{3}}e^{-2i\pi/3}}\cdot \frac{1}{\sqrt{3}}e^{-2i\pi/3}\right) + \left(\overline{\frac{1}{\sqrt{3}}e^{2i\pi/3}}\cdot\frac{1}{\sqrt{3}}e^{2i\pi/3}\right) \\ &= \frac 13 + \frac 13e^{+2i\pi/3}e^{-2i\pi/3} + \frac 13e^{-2i\pi/3}e^{2i\pi/3} \\ &= \frac 13 + \frac 13 e^0 + \frac 13 e^0 \\ &= \frac 13 + \frac 13 + \frac 13 \\ &= 1\end{align}$$