Illustration With Modulus 5

Let $p^{α} = 5$ .

$\begin{aligned} 2^{0} & \equiv 1 (\mod 5), & b (1) = {ind}_{2} 1 & = 0, \\ 2^{1} & \equiv 2 (\mod 5), & b (2) = {ind}_{2} 2 & = 1, \\ 2^{2} & \equiv 4 (\mod 5), & b (4) = {ind}_{2} 4 & = 2, \\ 2^{3} & \equiv 3 (\mod 5), & b (3) = {ind}_{2} 3 & = 3. \end{aligned}$

$χ_{h} (n) = {\begin{cases} e^{i 2 π h b (n) / 4} & if 5 ∤ n, \\ 0 & if 5 ∣ n . \end{cases}$

$\begin{array}{rrrrrr} n & 1 & 2 & 3 & 4 & 5 \\ χ_{0} (n) & e^{i 2 π \cdot 0 \cdot b (1) / 4} & e^{i 2 π \cdot 0 \cdot b (2) / 4} & e^{i 2 π \cdot 0 \cdot b (3) / 4} & e^{i 2 π \cdot 0 \cdot b (4) / 4} & 0 \\ χ_{1} (n) & e^{i 2 π \cdot 1 \cdot b (1) / 4} & e^{i 2 π \cdot 1 \cdot b (2) / 4} & e^{i 2 π \cdot 1 \cdot b (3) / 4} & e^{i 2 π \cdot 1 \cdot b (4) / 4} & 0 \\ χ_{2} (n) & e^{i 2 π \cdot 2 \cdot b (1) / 4} & e^{i 2 π \cdot 2 \cdot b (2) / 4} & e^{i 2 π \cdot 2 \cdot b (3) / 4} & e^{i 2 π \cdot 2 \cdot b (4) / 4} & 0 \\ χ_{3} (n) & e^{i 2 π \cdot 3 \cdot b (1) / 4} & e^{i 2 π \cdot 3 \cdot b (2) / 4} & e^{i 2 π \cdot 3 \cdot b (3) / 4} & e^{i 2 π \cdot 3 \cdot b (4) / 4} & 0 \end{array}$

$\begin{array}{rrrrrr} n & 1 & 2 & 3 & 4 & 5 \\ χ_{0} (n) & e^{i 2 π \cdot 0 \cdot 0 / 4} & e^{i 2 π \cdot 0 \cdot 1 / 4} & e^{i 2 π \cdot 0 \cdot 3 / 4} & e^{i 2 π \cdot 0 \cdot 2 / 4} & 0 \\ χ_{1} (n) & e^{i 2 π \cdot 1 \cdot 0 / 4} & e^{i 2 π \cdot 1 \cdot 1 / 4} & e^{i 2 π \cdot 1 \cdot 3 / 4} & e^{i 2 π \cdot 1 \cdot 2 / 4} & 0 \\ χ_{2} (n) & e^{i 2 π \cdot 2 \cdot 0 / 4} & e^{i 2 π \cdot 2 \cdot 1 / 4} & e^{i 2 π \cdot 2 \cdot 3 / 4} & e^{i 2 π \cdot 2 \cdot 2 / 4} & 0 \\ χ_{3} (n) & e^{i 2 π \cdot 3 \cdot 0 / 4} & e^{i 2 π \cdot 3 \cdot 1 / 4} & e^{i 2 π \cdot 3 \cdot 3 / 4} & e^{i 2 π \cdot 3 \cdot 2 / 4} & 0 \end{array}$

$\begin{array}{rrrrrr} n & 1 & 2 & 3 & 4 & 5 \\ χ_{0} (n) & 1 & 1 & 1 & 1 & 0 \\ χ_{1} (n) & 1 & i & - i & - 1 & 0 \\ χ_{2} (n) & 1 & - 1 & - 1 & 1 & 0 \\ χ_{3} (n) & 1 & - i & i & - 1 & 0 \end{array}$