Find norm of a function in $L_2(-1,1)$ (with complex numbers)

Question

I have the two functions $g_0(x):=(1+i)\exp(i \pi x)$ and $g_1(x):=\exp(i 2 \pi x)$ in $L_2(-1,1)$ so with inner product. I have to find $||g_0||$ and $||g_1||$. I think we have that $\left \| f \right \|=\sqrt{\left \langle f,f \right \rangle}=\sqrt{\int_{a}^{b}|f|^{2}dt}$. So I have to solve the integrals: $$\left \| g_0 \right \|=\sqrt{\left \langle g_0,g_0 \right \rangle}=\sqrt{\int_{a}^{b}|g_0|^{2}dx}=\sqrt{\int_{-1}^{1}|((1+i)\exp(i \pi x))|^{2}dx}$$ And $$\left \| g_1 \right \|=\sqrt{\left \langle g_1,g_1 \right \rangle}=\sqrt{\int_{a}^{b}|g_1|^{2}dx}=\sqrt{\int_{-1}^{1}|(\exp(i 2 \pi x))|^{2}dx}$$

But I'm not so good with integrals with complex numbers, can anyone help me? But I think that $||g_0||=2$ and $||g_1||=\sqrt{2}$ is my CAS have solved it correctly. Can anyone help me to show that?

Answer 1

Recall that $|\exp (i \pi x)| = |\exp (i 2\pi x)| = 1$ for all $x$. Thus, $$\|g_0\| = \sqrt{\int_{-1}^1 |(1+i)\exp (i \pi x)|^2dx} =\sqrt{\int_{-1}^1 |(1+i)|^2dx} = \sqrt{\int_{-1}^1 2 dx} = 2 $$

and

$$\|g_1\| = \sqrt{\int_{-1}^1 |\exp (i \pi x)|^2dx}=\sqrt{\int_{-1}^1 1dx} = \sqrt{2}$$