Understanding ( frac{1}{i} -i ): Simplifying Complex Numbers

Complex numbers play a vital role in many areas of mathematics and engineering. One of the simplest yet often confusing expressions is ( frac{1}{i} -i ). This article will break down the correct way to compute this expression and explain why a common incorrect logic is flawed.

Correct Calculation of ( frac{1}{i} )

The imaginary unit ( i ) is defined by the equation ( i^2 -1 ). Therefore, to simplify ( frac{1}{i} ) correctly, we can multiply the numerator and denominator by the complex conjugate of ( i ), which is ( -i ).

Step-by-Step Calculation

First, we rewrite the expression:

[ frac{1}{i} frac{1}{i} cdot frac{-i}{-i} ]

Next, we perform the multiplication in the denominator:

[ frac{-i}{i cdot -i} ]

Since (i^2 -1), we simplify the denominator:

[ frac{-i}{-(-1)} frac{-i}{1} -i ]

Why the Logic ( frac{1}{i} frac{1}{sqrt{-1}} ) is Incorrect

The common mistake arises when one tries to manipulate the expression as follows:

Breaking Down the Incorrect Logic

Start with the expression:

[ frac{1}{i} frac{1}{sqrt{-1}} ]

Then incorrectly split the square root:

[ frac{sqrt{1}}{sqrt{-1}} sqrt{frac{1}{-1}} sqrt{-1} ]

This is a flawed manipulation for the following reasons:

The property (sqrt{a} cdot sqrt{b} sqrt{ab}) only holds for non-negative ( a ) and ( b ). Here, (-1) is negative, making this property invalid.

The square root of a negative number, (sqrt{-1}), is not a real number but rather a complex number, which is (i).

Misinterpreting the properties of square roots of negative numbers can lead to incorrect results.

Therefore, the correct statement is that (frac{1}{i} -i), as demonstrated in the earlier calculation.

Summary

In summary, the correct calculation of (frac{1}{i}) is (-i) by using the complex conjugate method. The logic involving the manipulation of square roots is incorrect because it does not respect the properties of square roots in complex analysis.

Additional Insights: Euler's Formula

For a deeper understanding of complex numbers, consider Euler's formula:

[ e^{itheta} cos(theta) isin(theta) ]

When (theta pi/2), we get:

[ e^{ipi/2} cos(pi/2) isin(pi/2) i ]

This confirms that ( i e^{ipi/2} ), which further supports the property that (i^2 -1).