Key Concepts and Formulas

• Complex Conjugate: The complex conjugate of $a + bi$ is $a - bi$. Multiplying a complex number by its conjugate results in a real number: $(a + bi)(a - bi) = a^2 + b^2$.
• Powers of i: The powers of the imaginary unit $i$ cycle through four values: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. Thus, $i^k = 1$ if and only if $k$ is a multiple of 4, i.e., $k = 4p$ for some integer $p$.
• Greatest Common Divisor (GCD): The largest positive integer that divides two or more integers without a remainder.

Step-by-Step Solution

Step 1: Simplify the Base Expression

We simplify the expression $\frac{1+i}{1-i}$ by multiplying the numerator and denominator by the complex conjugate of the denominator, which is $1+i$. This eliminates the imaginary part from the denominator.

$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)}$

Expanding the numerator and denominator:

• Numerator: $(1+i)(1+i) = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$
• Denominator: $(1-i)(1+i) = 1 - i^2 = 1 - (-1) = 2$

Substituting back into the fraction:

$\frac{2i}{2} = i$

Why this step? Simplifying the base is crucial for making the problem tractable. It transforms a complex fraction into the simple imaginary unit, $i$.

Step 2: Substitute the Simplified Base into the Original Equation

Now that we have $\frac{1+i}{1-i} = i$, we substitute this into the given equation:

${\left( {{{1 + i} \over {1 - i}}} \right)^{{m \over 2}}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{{n \over 3}}} = 1$

becomes

$(i)^{m/2} = 1 \quad \text{and} \quad (i)^{n/3} = 1$

Why this step? This substitution simplifies the problem, allowing us to focus on the exponents of $i$.

Step 3: Solve for the Least Values of m and n

Since $i^k = 1$ if and only if $k$ is a multiple of 4, we require $\frac{m}{2}$ and $\frac{n}{3}$ to be multiples of 4.

For $m$: We have $\frac{m}{2} = 4k_1$, where $k_1$ is a positive integer (since $m \in N$). Solving for $m$:

$m = 8k_1$

To find the least value of $m$, we take $k_1 = 1$:

$m_{\text{least}} = 8(1) = 8$

For $n$: We have $\frac{n}{3} = 4k_2$, where $k_2$ is a positive integer. Solving for $n$:

$n = 12k_2$

To find the least value of $n$, we take $k_2 = 1$:

$n_{\text{least}} = 12(1) = 12$

Why this step? This step uses the cyclical property of $i$ to establish the conditions for $m$ and $n$. Finding the least values ensures we satisfy the problem's constraints.

Step 4: Calculate the Greatest Common Divisor (GCD)

We need to find the GCD of the least values of $m$ and $n$, which are 8 and 12, respectively.

Listing the divisors:

• Divisors of 8: {1, 2, 4, 8}
• Divisors of 12: {1, 2, 3, 4, 6, 12}

The greatest common divisor is 4.

Alternatively, using prime factorization:

• $8 = 2^3$
• $12 = 2^2 \times 3$

The GCD is the product of the lowest powers of their common prime factors: $2^2 = 4$.

$GCD(8, 12) = 4$

Why this step? This is the final calculation required by the problem statement.

Common Mistakes & Tips

• Tip: Memorize the simplification $\frac{1+i}{1-i} = i$ and $\frac{1-i}{1+i} = -i$ to save time.
• Mistake: Forgetting that $m$ and $n$ are natural numbers, so the constants $k_1$ and $k_2$ must be positive integers (not zero).
• Mistake: Confusing the powers of $i$. Remember $i^4 = 1$, not $i^2 = 1$.

Summary

We simplified the complex fraction to $i$, then used the property that $i^k = 1$ when $k$ is a multiple of 4 to find the least values of $m$ and $n$. We found $m = 8$ and $n = 12$, and then calculated their greatest common divisor, which is 4.

The final answer is $\boxed{4}$.