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 \((m – n)\)?
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 \(i = \sqrt{-1}\).
To solve this, we first simplify the complex fractions that form the bases of the 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 \((-i)^2 = -1\).
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.
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.
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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