Simplifying Complex Number Expressions

The problem requires finding the smallest positive integer value for the expression $m – n$, given the specific equation involving complex numbers: $ \left( \frac{1-i}{1+i} \right)^{2m} \left( \frac{1+i}{1-i} \right)^{2n} = 1 $ Here, $i$ represents the imaginary unit, where $i = \sqrt{-1}$.

Simplifying Base Fractions

To solve this, we first simplify the complex fractions that form the bases of the powers.

• Simplifying $\frac{1-i}{1+i}$: We multiply the numerator and denominator by the complex conjugate of the denominator, which is $(1-i)$. $ \frac{1-i}{1+i} = \frac{(1-i) \times (1-i)}{(1+i) \times (1-i)} = \frac{1 - i - i + i^2}{1^2 - i^2} $ Using the fact that $i^2 = -1$: $ \frac{1 - 2i + (-1)}{1 - (-1)} = \frac{1 - 2i - 1}{1 + 1} = \frac{-2i}{2} = -i $
• Simplifying $\frac{1+i}{1-i}$: Similarly, we multiply the numerator and denominator by the complex conjugate of the denominator, which is $(1+i)$. $ \frac{1+i}{1-i} = \frac{(1+i) \times (1+i)}{(1-i) \times (1+i)} = \frac{1 + i + i + i^2}{1^2 - i^2} $ Using $i^2 = -1$: $ \frac{1 + 2i + (-1)}{1 - (-1)} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i $

Substituting and Simplifying Powers

Now, we substitute these simplified results, $-i$ and $i$, back into the original equation:

Let's simplify these powers. We can use the property $(a^b)^c = a^{bc}$ and the values $i^2 = -1$ and $(-i)^2 = -1$.

• $(-i)^{2m} = ((-i)^2)^m = (-1)^m$
• $(i)^{2n} = ((i)^2)^n = (-1)^n$

Substituting these simplified powers back into the equation gives:

For the equation $(-1)^{m+n} = 1$ to hold true, the exponent $(m+n)$ must be an even integer. We can express this condition as $m+n = 2k$, where $k$ is any integer.

Deriving the Condition on $(m-n)$

An alternative way to analyze the powers is by using the polar form of complex numbers. We know $i = e^{i\pi/2}$ and $-i = e^{-i\pi/2}$.

The equation becomes:

The complex exponential $e^{i\theta}$ equals 1 if and only if $\theta$ is an integer multiple of $2\pi$. Therefore, we must have:

where $k$ is an integer. Dividing both sides by $\pi$ gives:

This confirms that the difference $(n-m)$ must be an even integer.

Finding the Smallest Positive Value

Since $n-m$ must be an even integer, its negative, $m-n = -(n-m)$, must also be an even integer.

The set of all possible even integers is $\{\dots, -6, -4, -2, 0, 2, 4, 6, \dots\}$.

The question asks for the smallest positive value of $(m-n)$. From the set of possible even integer values, the smallest positive integer is 2.