Powers of i repeat every 4: i¹=i, i²=-1, i³=-i, i⁴=1. Find remainder of 47 ÷ 4. JEE favourite.

What is i^47? — Powers of i and the Cycle of Four

Question

Find the value of $i^{47}$ .

Solution — Step by Step

The imaginary unit $i$ follows a repeating cycle of length 4:

i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1

After every 4 powers, we’re back to 1. This is the key pattern the entire topic is built on.

We need the remainder when 47 is divided by 4:

47 = 4 \times 11 + 3

So the remainder is 3.

Since $47 = 4 \times 11 + 3$ , we can write:

i^{47} = i^{4 \times 11 + 3} = (i^4)^{11} \cdot i^3

Now $(i^4)^{11} = 1^{11} = 1$ , so:

i^{47} = 1 \cdot i^3 = i^3 = -i

Answer: $i^{47} = -i$

Why This Works

The pattern $i, -1, -i, 1$ repeats because each multiplication by $i$ rotates a point 90° anticlockwise in the Argand plane. After four 90° rotations, you’re back where you started — that’s a full 360°.

This is why $i^4 = 1$ is not a coincidence. It’s a geometric fact dressed in algebra. Every time the exponent is a multiple of 4, you land back at 1.

The division algorithm does the heavy lifting here. Any integer $n$ can be written as $n = 4q + r$ where $r \in \{0, 1, 2, 3\}$ . The value of $i^n$ depends only on $r$ — the “excess” after all the complete cycles cancel out.

The four remainders map directly:

Remainder 0 → $i^n = 1$
Remainder 1 → $i^n = i$
Remainder 2 → $i^n = -1$
Remainder 3 → $i^n = -i$

Memorise this table once. Every powers-of- $i$ question in JEE Main and CBSE Class 11 reduces to finding which remainder you’re in.

Alternative Method

We can also use negative exponents to simplify large powers. Notice that:

i^{47} = i^{48} \cdot i^{-1}

Since $48 = 4 \times 12$ , we have $i^{48} = 1$ . And $i^{-1} = \frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$ .

So $i^{47} = 1 \cdot (-i) = -i$ .

This approach is useful when the exponent is just 1 below a known multiple of 4 — spotting $48 = 4 \times 12$ is faster than doing the division algorithm. Both methods are equally valid for JEE.

Common Mistake

Dividing 47 by 2 instead of 4.

Students sometimes remember “i cycles” but forget the cycle length is 4, not 2. They compute $47 \div 2 = 23$ remainder 1, then write $i^{47} = i^1 = i$ . Wrong.

The cycle has length 4 because it takes four multiplications by $i$ to get back to 1. The remainder must always be taken mod 4. In JEE Main 2019, a similar question — $i^{-999}$ — was marked incorrectly by a large number of students for exactly this reason.

For $i^{-999}$ : remainder of $999 \div 4$ is $3$ , so $i^{-999} = i^{-3}$ . Then $i^{-3} = \frac{1}{i^3} = \frac{1}{-i} = \frac{i}{1} = i$ (after rationalising). Watch for negative exponents in JEE — they are a separate trap on top of the cycle rule.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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