Question

Given that $i^{2}=-1$. Then $(i^{1})(i^{2})(i^{3})\dots(i^{2026})$ is equal to

• A. -1
• B. 1
• C. 2i
• D. i
• E. -i

Hint:

Math Tip: When dealing with large powers of $i$, you only need to care about the remainder of the exponent when divided by 4.
$i^{4k} = 1$, \quad $i^{4k+1} = i$, \quad $i^{4k+2} = -1$, \quad $i^{4k+3} = -i$.

Answer 1

Answer: (E)

Concept:
Complex Numbers - Powers of $i$ and Arithmetic Progressions.
Step 1: Combine the terms using exponent rules.
When multiplying terms with the same base, you add their exponents: $$ (i^{1})(i^{2})(i^{3})\dots(i^{2026}) = i^{1 + 2 + 3 + \dots + 2026} $$
Step 2: Calculate the sum of the arithmetic progression in the exponent.
The exponent is the sum of the first $n$ natural numbers, where $n = 2026$.
Use the formula $S_n = \frac{n(n+1)}{2}$. $$ \text{Sum} = \frac{2026 \times (2026 + 1)}{2} $$ $$ \text{Sum} = \frac{2026 \times 2027}{2} = 1013 \times 2027 $$
Step 3: Evaluate the exponent modulo 4.
The powers of $i$ repeat in a cycle of 4: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.
To find the final value, we need to find the remainder when the exponent is divided by 4.
Let's find $(1013 \times 2027) \pmod 4$. $$ 1013 \pmod 4 \equiv 1 \text{ (since } 1012 \text{ is divisible by 4)} $$ $$ 2027 \pmod 4 \equiv 3 \text{ (since } 2024 \text{ is divisible by 4)} $$
Step 4: Determine the final power of i.
Multiply the remainders to find the combined modulo: $$ (1 \times 3) \pmod 4 = 3 $$ This means the large exponent is of the form $4k + 3$.
Therefore: $$ i^{1013 \times 2027} = i^{4k+3} = i^3 $$
Step 5: State the final result.
Since $i^3 = i^2 \times i = (-1) \times i = -i$, the entire product evaluates to $-i$.