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Step-by-Step Solution
Step 1: Understand the Given Expression
The problem states that:
\left(\frac{1 + i}{1 - i}\right)^{m/2} \;=\; \left(\frac{1 + i}{1 - i}\right)^{n/3} \;=\; 1
where m,n \in \mathbb{N} . We need to find the greatest common divisor (GCD) of the least values of m and n that satisfy this equality.
Step 2: Simplify the Base \left(\frac{1 + i}{1 - i}\right)
First, we notice:
(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i,
and
(1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i.
Therefore,
\frac{1 + i}{1 - i}
= \frac{(1 + i)^2}{(1 + i)(1 - i)}
= \frac{2i}{1 - i^2}
= \frac{2i}{1 - (-1)}
= \frac{2i}{2} = i.
However, in many treatments, one might directly see that \frac{1 + i}{1 - i} = i , but let's follow the problem’s own hint-like approach. The expression becomes:
\left(\frac{1 + i}{1 - i}\right)^{m/2}
= i^{m/2}
\quad \text{and} \quad
\left(\frac{1 + i}{1 - i}\right)^{n/3}
= i^{n/3}.
According to the question, each equals 1.
Step 3: Use the Fact i^4 = 1
We know powers of i cycle every 4 steps:
i^1 = i, \quad
i^2 = -1, \quad
i^3 = -i, \quad
i^4 = 1.
If i^{m/2} = 1 , this implies:
\frac{m}{2} \equiv 0 \pmod{4},
or m/2 = 4k_1 for some integer k_1 . Hence:
m = 8k_1.
Similarly, for i^{n/3} = 1 , we need:
\frac{n}{3} \equiv 0 \pmod{4},
or n/3 = 4k_2 for some integer k_2 . Hence:
n = 12k_2.
The least positive value of m occurs when k_1 = 1 , giving:
m = 8.
and the least positive value of n occurs when k_2 = 1 , giving:
n = 12.
Step 4: Compute the GCD of the Least Values
The least values of m and n are 8 and 12, respectively. Their greatest common divisor is:
\gcd(8, 12) = 4.
Final Answer
The greatest common divisor of the least values of m and n is 4.
