JEE Main 2025 — 28 January, Morning Shift. Previous Year Question.
Problem
Two numbers $k_1$ and $k_2$ are randomly chosen from the set of natural numbers. Then the probability that the value of $i^{k_1}+i^{k_2}$ (where $i=\sqrt{-1}$) is non-zero, is equal to:
Key insight. $i^k$ only ever takes one of four values, and it repeats with period 4 — so the whole problem reduces to picking two values, each uniformly at random from $\{i,-1,-i,1\}$, and asking how often they don’t cancel.
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Approach
Since $i^k$ depends only on $k \bmod 4$, treat $k_1 \bmod 4$ and $k_2 \bmod 4$ as independent and uniform over the four residues $1,2,3,0$ (equivalently $4$). List the 16 equally likely combinations, count how many make $i^{k_1}+i^{k_2}=0$, and subtract that fraction from 1.
Solution
Step 1 — The four possible values of $i^k$
$$i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1$$
and this block of four repeats forever as $k$ increases, since $i^4=1$. So whichever natural number $k$ is chosen, $i^k$ is always one of $\{i,-1,-i,1\}$, each occurring for exactly one residue class of $k$ modulo 4.
Step 2 — When does the sum vanish?
$i^{k_1}+i^{k_2}=0$ exactly when $i^{k_2}=-i^{k_1}$, i.e. the two values are negatives of each other. Among $\{i,-1,-i,1\}$, the only such pairs are $(i,-i)$ and $(1,-1)$ — each element has exactly one “partner” that cancels it. In terms of residues mod 4, the zero-sum pairs of $(k_1 \bmod 4,\, k_2 \bmod 4)$ are
$$(1,3),\ (3,1),\ (2,4),\ (4,2)$$
Step 3 — Count over all equally likely outcomes
Since $k_1 \bmod 4$ and $k_2 \bmod 4$ each take one of 4 values independently, there are $4\times4=16$ equally likely combinations in total. Exactly 4 of them are zero-sum, so
$$P(i^{k_1}+i^{k_2}=0)=\frac{4}{16}=\frac14$$
Step 4 — Take the complement
The event asked for is the opposite one — that the sum is non-zero:
$$P(i^{k_1}+i^{k_2}\neq0)=1-\frac14=\frac34$$
Answer
$$\frac34$$
Common mistakes
- Treating all 16 combinations as if they need to be listed to see the pattern, then miscounting the cancelling pairs. It helps to notice the cycle has exactly two “opposite” pairs, $\{i,-i\}$ and $\{1,-1\}$, and each ordered pair contributes twice (once each way round) — that’s where the 4 out of 16 comes from.
- Forgetting the complement step. The question asks for the probability of a non-zero sum, not a zero sum — it’s easy to correctly compute $\frac14$ and then submit that as the final answer instead of $1-\frac14$.
Practise next
- Two numbers $m$ and $n$ are randomly chosen from the natural numbers. Find the probability that $i^m \cdot i^n = -1$, using the same residue-mod-4 approach.
Show answer
$\dfrac14$. $i^m\cdot i^n=i^{m+n}=-1$ exactly when $m+n\equiv2\pmod4$.
‘Randomly chosen from the natural numbers’ is read, as usual, as each residue modulo $4$ being equally likely. Of the $16$ residue pairs, four give a sum $\equiv2$: $(0,2)$, $(2,0)$, $(1,1)$, $(3,3)$.
So the probability is $\tfrac4{16}=\tfrac14$. (There is no uniform distribution on all of $\mathbb N$, which is why the residue reading is the one that makes sense.)
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.