Question

Let $Z_1, Z_2, Z_3$ and $Z_4$ be four complex numbers chosen independently from the set $S = \{1, -1, i, -i\}$. If $Z = Z_1 + Z_2 + Z_3 + Z_4$. Then the probability that $Z$ is a real number and $|Z| = 2$ is (where $i = \sqrt{-1}$)

• A. 3/4
• B. 1/4
• C. 1/3
• D. 1/8

Answer 1

Answer: (D)

1/8

Answer 2

Let $n_1, n_{-1}, n_i, n_{-i}$ be the counts of $1, -1, i, -i$ respectively, such that $n_1+n_{-1}+n_i+n_{-i}=4$. The sum is $Z = (n_1-n_{-1}) + i(n_i-n_{-i})$. For $Z$ to be real, $n_i=n_{-i}$. For $|Z|=2$, we need $|n_1-n_{-1}|=2$.

1. If $n_i=n_{-i}=0$, then $n_1+n_{-1}=4$. $|n_1-n_{-1}|=2$ yields $(n_1,n_{-1})=(3,1)$ or $(1,3)$. Number of ways: $\binom{4}{3,1,0,0} + \binom{4}{1,3,0,0} = 4+4=8$.
2. If $n_i=n_{-i}=1$, then $n_1+n_{-1}=2$. $|n_1-n_{-1}|=2$ yields $(n_1,n_{-1})=(2,0)$ or $(0,2)$. Number of ways: $\binom{4}{2,0,1,1} + \binom{4}{0,2,1,1} = 12+12=24$.
3. If $n_i=n_{-i}=2$, then $n_1+n_{-1}=0$, so $(n_1,n_{-1})=(0,0)$. Sum is $Z=0$, so $|Z|=0 \ne 2$. Number of ways = 0.

Total favorable ways = $8+24=32$. Total possible ways = $4^4=256$. Probability = $32/256 = 1/8$.