If \(\left( \frac{1-i}{1+i} \right)^{2m} \left( \frac{1+i}{1-i} \right)^{2n} = 1\) where \(i = \sqrt{-1}\), then what is the smallest positive value of \((m – n)\)?
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:
\((-i)^{2m} (i)^{2n} = 1\)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:
\((-1)^m \times (-1)^n = 1\) \((-1)^{m+n} = 1\)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:
\((e^{-i\pi/2})^{2m} \times (e^{i\pi/2})^{2n} = 1\) \(e^{-i\pi m} \times e^{i\pi n} = 1\) \(e^{i\pi (n-m)} = 1\)The complex exponential \(e^{i\theta}\) equals 1 if and only if \(\theta\) is an integer multiple of \(2\pi\). Therefore, we must have:
\(\pi(n-m) = 2k\pi\)where \(k\) is an integer. Dividing both sides by \(\pi\) gives:
\(n-m = 2k\)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.
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