Imaginary unit i = √(−1), i² = −1, powers of i cycle: Exercises

Here is the compass picture we will lean on for the geometric problems.

Figure — Imaginary unit i = √(−1), i² = −1, powers of i cycle

Level 1 — Recognition

Goal: read off the value directly from the 4-cycle, no arithmetic tricks.

L1.1 Write down $i^{1}, i^{2}, i^{3}, i^{4}$ from memory.

Recall Solution L1.1

What we do: just recall the cycle. Why: these four are the atoms everything else reduces to. $i^{1} = i, i^{2} = - 1, i^{3} = - i, i^{4} = 1.$ Picture: starting at $1$ (east), each step turns $9 0^{\circ}$ anticlockwise → north → west → south → back to east.

L1.2 What is $i^{0}$ , and why?

Recall Solution L1.2

What we do: apply the ordinary exponent rule "anything nonzero to the power $0$ is $1$ ." Why it holds for $i$ : $i \neq = 0$ , so the rule applies unchanged. Also $i^{0} = i^{4} = 1$ agrees with the cycle. $i^{0} = 1.$

L1.3 True or false: $i^{3} = i$ . Justify in one line.

Recall Solution L1.3

False. $i^{3} = i^{2} \cdot i = (- 1) \cdot i = - i$ . The mistake is forgetting to collapse $i \cdot i = - 1$ . $i^{3} = - i .$

Level 2 — Application

Goal: use $r = n mod 4$ on a single power.

L2.1 Compute $i^{22}$ .

Recall Solution L2.1

Step 1 — divide by 4: $22 = 4 \times 5 + 2$ , so the remainder is $r = 2$ . Why: only the leftover after removing full loops of 4 decides the landing spot. Step 2 — read the cycle: $i^{22} = i^{2} = - 1.$ $i^{22} = - 1.$

L2.2 Compute $i^{45}$ .

Recall Solution L2.2

Step 1: $45 = 4 \times 11 + 1$ , so $r = 1$ . Step 2: $i^{45} = i^{1} = i .$ $i^{45} = i .$

L2.3 Compute $i^{- 1}$ and $i^{- 3}$ .

Recall Solution L2.3

$i^{- 1}$ : multiply top and bottom by $i$ to clear $i$ from the denominator (like rationalising): $i^{- 1} = \frac{1}{i} = \frac{1}{i} \cdot \frac{i}{i} = \frac{i}{i ^{2}} = \frac{i}{- 1} = - i .$ $i^{- 3}$ : the cycle also runs backwards with period 4. Add 4 to the exponent until it is in ${0, 1, 2, 3}$ : $- 3 + 4 = 1$ , so $i^{- 3} = i^{1} = i$ . $i^{- 1} = - i, i^{- 3} = i .$

Level 3 — Analysis

Goal: combine several powers and exploit structure.

L3.1 Simplify $i^{12} + i^{13} + i^{14} + i^{15}$ .

Recall Solution L3.1

Step 1 — reduce each: $12 \to r = 0 \Rightarrow 1$ ; $13 \to r = 1 \Rightarrow i$ ; $14 \to r = 2 \Rightarrow - 1$ ; $15 \to r = 3 \Rightarrow - i$ . Step 2 — add: $1 + i + (- 1) + (- i) = 0.$ Why zero: any four consecutive powers are ${1, i, - 1, - i}$ in some order, and these sum to $0$ . $i^{12} + i^{13} + i^{14} + i^{15} = 0.$

L3.2 Compute $i^{1} + i^{2} + i^{3} + \dots + i^{20}$ (twenty terms).

Recall Solution L3.2

Step 1 — group in fours: $20$ terms make exactly $5$ blocks of $4$ consecutive powers: $(i^{1} + i^{2} + i^{3} + i^{4}) + \dots$ Step 2 — each block is $0$ : so the total is $5 \times 0 = 0$ . Why grouping works: the cycle repeats every 4, so each block is a shuffled ${1, i, - 1, - i}$ . $\sum_{n = 1}^{20} i^{n} = 0.$

L3.3 Compute $i^{1} + i^{2} + i^{3} + \dots + i^{23}$ (twenty-three terms).

Recall Solution L3.3

Step 1 — full blocks: $23 = 4 \times 5 + 3$ . The first $20$ terms give $5$ complete blocks $= 0$ . Step 2 — leftover 3 terms are $i^{21} + i^{22} + i^{23}$ . Reduce: $21 \to r = 1 \Rightarrow i$ ; $22 \to r = 2 \Rightarrow - 1$ ; $23 \to r = 3 \Rightarrow - i$ . Step 3 — add leftovers: $i + (- 1) + (- i) = - 1.$ $\sum_{n = 1}^{23} i^{n} = - 1.$

Level 4 — Synthesis

Goal: mix reciprocals, square-root traps, and the cycle.

L4.1 Evaluate $\frac{i ^{18}}{i ^{7}}$ .

Recall Solution L4.1

Step 1 — subtract exponents: $\frac{i ^{18}}{i ^{7}} = i^{18 - 7} = i^{11}$ . Why: dividing same bases subtracts exponents, and this is legal since $i \neq = 0$ . Step 2 — reduce: $11 = 4 \times 2 + 3$ , so $i^{11} = i^{3} = - i .$ $\frac{i ^{18}}{i ^{7}} = - i .$

L4.2 Compute $- 16 \cdot - 25$ .

Recall Solution L4.2

Step 1 — convert to $i$ FIRST: $- 16 = 4 i$ and $- 25 = 5 i$ . Why first: the identity $a b = ab$ is only valid for non-negative reals, so we must not merge two negatives under one root. Step 2 — multiply: $(4 i) (5 i) = 20 i^{2} = 20 (- 1) = - 20.$ $- 16 \cdot - 25 = - 20.$

L4.3 Simplify $(1 + i)^{2}$ .

Recall Solution L4.3

Step 1 — expand like any binomial: $(1 + i)^{2} = 1^{2} + 2 \cdot 1 \cdot i + i^{2} = 1 + 2 i + i^{2}$ . Step 2 — substitute $i^{2} = - 1$ : $1 + 2 i - 1 = 2 i .$ $(1 + i)^{2} = 2 i .$ Bonus insight: $1 + i$ points $4 5^{\circ}$ from east; squaring doubles the angle to $9 0^{\circ}$ (straight up), which is exactly the direction of $2 i$ . Geometry agrees with algebra.

Level 5 — Mastery

Goal: prove a general fact and reason geometrically.

L5.1 (Proof) Show that for every integer $k \geq 0$ , $i^{k} + i^{k + 1} + i^{k + 2} + i^{k + 3} = 0.$

Recall Solution L5.1

Step 1 — factor out the common power: pull $i^{k}$ from every term: $i^{k} + i^{k + 1} + i^{k + 2} + i^{k + 3} = i^{k} (1 + i + i^{2} + i^{3}) .$ Why: this isolates the part that does not depend on $k$ , so we only need to evaluate the bracket once. Step 2 — evaluate the bracket using $i^{2} = - 1, i^{3} = - i$ : $1 + i + i^{2} + i^{3} = 1 + i - 1 - i = 0.$ Step 3 — conclude: $i^{k} \cdot 0 = 0$ for any $k$ . ∎ This is why any four consecutive powers vanish, regardless of where you start — the $i^{k}$ factor never matters because the bracket is exactly $0$ .

L5.2 (Geometry) Using the rotation meaning of $i$ , explain without arithmetic why $i^{4} = 1$ , then predict $i^{6}$ .

Recall Solution L5.2

What multiplying by $i$ does: it rotates a point $9 0^{\circ}$ anticlockwise about the origin (see the compass figure). Why $i^{4} = 1$ : starting at $1$ (east), four turns of $9 0^{\circ}$ total $36 0^{\circ}$ — a full circle — landing back exactly on $1$ . So $i^{4} = 1$ with no computation. Predict $i^{6}$ : six turns $= 6 \times 9 0^{\circ} = 54 0^{\circ}$ . Subtract one full loop ( $36 0^{\circ}$ ) to get $18 0^{\circ}$ , which points due west — the location of $- 1$ . Hence $i^{6} = - 1$ , matching $6 mod 4 = 2 \Rightarrow i^{2} = - 1$ .

L5.3 (Synthesis) Evaluate $i^{2024} + i^{2025} + i^{2026}$ .

Recall Solution L5.3

Step 1 — reduce each: $2024 = 4 \times 506 \Rightarrow r = 0 \Rightarrow 1$ ; $2025 \Rightarrow r = 1 \Rightarrow i$ ; $2026 \Rightarrow r = 2 \Rightarrow - 1$ . Step 2 — add: $1 + i + (- 1) = i .$ $i^{2024} + i^{2025} + i^{2026} = i .$ Sanity check: these are three of a four-block; a full block sums to $0$ , so three terms equal $- (missing i^{2027}) = - (i^{3}) = - (- i) = i$ . ✓

Connections

Complex Numbers — parent framework $a + bi$ these exercises live in.
Hinglish version of the topic.
Argand Diagram — the compass picture behind L5.2.
Modulus and Argument — squaring $1 + i$ doubling its angle (L4.3).
Roots of Unity — ${1, i, - 1, - i}$ are the four 4th roots of unity that make the blocks sum to $0$ .
Euler's Formula — $i = e^{iπ /2}$ turns "multiply by $i$ " into "add $9 0^{\circ}$ ."