Powers of i

, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, $i^{4n} = 1$, $i^{4n+1} = i$, $i^{4n+2} = -1$, $i^{4n+3} = -i$

Deep Breakdown

What it tells me (Situation Awareness): The powers of $i$ cycle with period 4: $\{i, -1, -i, 1, i, -1, -i, 1, \ldots\}$. This is a direct consequence of Euler's formula: $i = e^{i\pi/2}$, so $i^n = e^{in\pi/2}$, which completes a full rotation ($2\pi$) every 4 steps. In DA/DS, the cyclic pattern of $i$-powers is the algebraic foundation for: simplifying complex arithmetic, understanding why the DFT has period $N$, and working with roots of unity. It is also the pattern you use every time you simplify $i^{35}$ (without a calculator).

Flowchart

Real-world use

Where it is used

Simplifying complex expressions: $i^{43} = i^{4 \times 10 + 3} = (i^4)^{10} \cdot i^3 = 1^{10} \cdot (-i) = -i$. Roots of unity: the 4th roots of unity are $1, i, -1, -i$ (the four powers of $i$). FFT butterfly: $i^k$ for $k = 0, 1, 2, 3$ are the "twiddle factors" of the length-4 DFT. Signal rotation: multiplying by $i$ rotates by 90°, multiplying by $i^2 = -1$ rotates by 180°.

Problem it solves

When computing large powers of $i$, the cyclic pattern gives the answer instantly: compute $n \bmod 4$ and read the table. Without the cyclic pattern, you'd need to expand all the multiplications. With it: $i^{1000} = i^{4 \times 250} = (i^4)^{250} = 1$.

How it solves the problem

Since $i^4 = 1$, any power $i^n = i^{n \bmod 4}$.

The four values are:

• $n \bmod 4 = 0$: $i^n = 1$
• $n \bmod 4 = 1$: $i^n = i$
• $n \bmod 4 = 2$: $i^n = -1$
• $n \bmod 4 = 3$: $i^n = -i$ This is a finite cyclic group $\mathbb{Z}_4 = \{0, 1, 2, 3\}$ under modular arithmetic, and $\{1, i, -1, -i\}$ is the corresponding group of 4th roots of unity.

Example in a business

DFT twiddle factors for $N = 4$: the matrix is built from $W_4^{jk} = e^{-i2\pi jk/4} = e^{-i\pi jk/2} = (-i)^{jk}$. For $j, k \in \{0,1,2,3\}$, the twiddle factors are powers of $-i$, which follows the same 4-cycle. FFT algorithms rely on the fact that these twiddle factors cycle with period 4 (or $N$ generally), enabling the divide-and-conquer recursion.

The formula, piece by piece

Breaking the notation:

• $i^1 = i$: $i$ to the first power is $i$.
• $i^2 = -1$: the defining property ($i^2 = -1$, definition of $i$).
• $i^3 = i^2 \cdot i = -1 \cdot i = -i$.
• $i^4 = i^3 \cdot i = (-i) \cdot i = -i^2 = -(-1) = 1$.
• $i^{4n} = (i^4)^n = 1^n = 1$.
• $i^{4n+1} = i^{4n} \cdot i = 1 \cdot i = i$.
• $i^{4n+2} = i^{4n} \cdot i^2 = 1 \cdot (-1) = -1$.
• $i^{4n+3} = i^{4n} \cdot i^3 = 1 \cdot (-i) = -i$. Compact rule: $i^n = i^{n \bmod 4}$, with the table $\{0 \to 1,\ 1 \to i,\ 2 \to -1,\ 3 \to -i\}$.

Negative powers:

• $i^{-1} = 1/i = i/i^2 = i/(-1) = -i = i^3$.
• $i^{-2} = (i^2)^{-1} = (-1)^{-1} = -1 = i^2$.
• In general: $i^{-n} = i^{4-n \bmod 4}$ (just go backwards in the cycle).

The formula in words

"The imaginary unit cycles through four values when you raise it to successive powers: $i$, $-1$, $-i$, $1$, then back to $i$. It's like a clock with four positions. To find any power of $i$: divide the exponent by 4, look at the remainder, and look up the result in the four-entry table."

Critical Concepts

1. Quick computation rule: n % 4 gives the remainder. Map $0 \to 1$, $1 \to i$, $2 \to -1$, $3 \to -i$.
2. 4th roots of unity: $\{1, i, -1, -i\}$ are exactly the four 4th roots of unity: each when raised to the 4th power equals 1. These form a cyclic group of order 4.
3. Unit circle connection: $i = e^{i\pi/2}$ sits at 90° on the unit circle. Multiplying by $i$ is a 90° counter-clockwise rotation. Four such rotations complete a full circle and return to the original point.
4. Python: (1j)**n works for integer $n$.

Note that Python returns exact complex results for small integer powers of 1j.

Step-by-step calculation (tiny example)

Task: find $i^{43}$ and $i^{-7}$.

Step 1: $i^{43}$ $43 \bmod 4 = 43 - 4 \times 10 = 43 - 40 = 3$. $i^{43} = i^3 = -i$.

Step 2: Verify by factoring $i^{43} = i^{40} \cdot i^3 = (i^4)^{10} \cdot (-i) = 1^{10} \cdot (-i) = -i$. Confirmed.

Step 3: $i^{-7}$ $-7 \bmod 4$: $-7 = 4 \times (-2) + 1$, so $-7 \equiv 1 \pmod 4$. $i^{-7} = i^1 = i$. Alternatively: $i^{-7} = i^{-8} \cdot i = (i^4)^{-2} \cdot i = 1 \cdot i = i$. Confirmed.

Code examples

Beginner example (Python)

# Powers of i: cycle of length 4
# i^1=i, i^2=-1, i^3=-i, i^4=1, then repeats
import numpy as np   # Compute and display first 12 powers of i
print("Powers of i (1j in Python):")
print(f"{'n':>4} {'i^n':>10} {'n mod 4':>8} {'formula':>8}")
for n in range(1, 13):
    power = (1j)**n
remainder = n % 4
formula = {0: "1", 1: "i", 2: "-1", 3: "-i[["}[remainder]
print(f"]] {n:>2} {str(power):>10} {remainder:>8} {formula:>8}")   # Fast rule: i^n = i^(n mod 4)

def power_of_i(n: int) -> complex:
    """Fast computation of i^n using cyclic property."""
    lookup = {0: 1+0j, 1: 0+1j, 2: -1+0j, 3: 0-1j}
    return lookup[n % 4]   # Test large powers
test_powers = [43, 100, -7, 1001, 0]
print("\nLarge power lookup:")
for n in test_powers:
    fast = power_of_i(n)
direct = (1j)**n
print(f" i^{n:>5} = {fast:>7} (direct: {direct:>7}) match: {abs(fast-direct)<1e-10}")   # DFT 4-point twiddle factors W = e^(-i*2π/N) for
N=4
N = 4
W = np.exp(-1j * 2 * np.pi / N)   # = e^(-iπ/2) = -i
print(f"\n4-point DFT twiddle base: W_4 = e^(-iπ/2) = {W:.4f}")
print("Twiddle factors W_4^k:")
for k in range(N):
    wk = W**k
print(f" W_4^{k} = {wk:.4f} = i^{(-k)%4} = {power_of_i(-k):.4f}")

Beginner example (R)

# Powers of i in R
power_of_i <- function(n) { lookup <- list("0"=1+0i, "1"=0+1i, "2"=-1+0i, "3"=0-1i)
  key <- as.character(n %% 4)
  return(lookupkey)
}
cat("Powers of i:\n")
for (n in 1:8) { p <- (1i)^n
  f <- power_of_i(n)
  cat(sprintf(" i^%d = %s (n%%4=%d, fast=%s)\n", n, p, n%%4, f))
}
cat("\nLarge powers:\n")
for (n in c(43, 100, -7, 1001)) { fast <- power_of_i(n)
  cat(sprintf(" i^%d = %s\n", n, fast))
}

Professional example (Python)

import numpy as np   # Scenario: 4-point DFT matrix built from powers of i
# Demonstrates cyclic structure in Fourier analysis
N = 4
print("4-Point DFT Matrix (DFT_4) built from powers of i")
print("=" * 50)
print(" DFT_4[j,k] = e^(-i·2π·j·k/4) = (e^(-iπ/2))^(j·k) = (-i)^(j·k)")
print()   # Build DFT matrix
W_base = np.exp(-1j * 2 * np.pi / N)   # = -i (base twiddle)
DFT_4 = np.array([[W_base**(j*k) for k in range(N)] for j in range(N)])
print(" DFT_4 matrix:")
for row in DFT_4:
    row_str = " [" + " ".join(f"{v.real:+.0f}{v.imag:+.0f}j" for v in row) + " ]"
print(row_str)
print()   # Verify: each entry is a power of (-i) = i^3 = i^{4n+3}
print(" Entries as powers of i (via j*k mod 4 lookup):")
lookup = {0: " 1", 1: " i", 2: "-1", 3: "-i"}
for j in range(N):
    row_strs = []
for k in range(N):
    exponent = (j * k) % N   # power of W = power of (-i) # W_base = e^(-iπ/2) = -i = i^3; so W^(j*k) = (-i)^(j*k)
row_strs.append(f"(-i)^{j*k}={lookup[(j*k*3)%4]}")
print(f" j={j}: [{', '.join(row_strs)}]")
print()
# Apply DFT to a simple signal and verify
signal = np.array([1.0, 0.0, 1.0, 0.0])   # alternating 1, 0, 1, 0
X_matrix = DFT_4 @ signal
X_fft = np.fft.fft(signal)
print("DFT of [1, 0, 1, 0] via matrix vs. np.fft.fft:")
for k, (m, f) in enumerate(zip(X_matrix, X_fft)):
    print(f" X[{k}]: matrix={m:.3f} fft={f:.3f} match: {abs(m-f)<1e-10}")

Professional example (R)

# DFT via matrix (powers of i in entries) vs fft()
signal <- c(1, 0, 1, 0)
N <- 4
k <- 0:(N - 1)
n <- 0:(N - 1)
W <- outer(k, n, function(k, n) exp(-2i * pi * k * n / N))
X_matrix <- as.vector(W %*% signal)
X_fft <- fft(signal)
cat("DFT of [1,0,1,0] via matrix vs fft():\n")
for (i in seq_along(X_matrix)) { cat(sprintf(" X[%d]: matrix=%.3f fft=%.3f match: %s\n", i-1, X_matrix[i], X_fft[i], abs(X_matrix[i] - X_fft[i]) < 1e-10))
}

Source: domain knowledge

#foundations#number-sets-and-logic#complex-numbers-foundations