Proving Complex Logarithmic Relationships Using Euler’s Formula

Understanding how to manipulate and prove complex logarithmic relationships can be challenging. This article guides you through the process using Euler's formula to simplify arguments and find a common ground for solving complex equations. We'll explore the proof of how to derive a specific logarithmic relationship involving square roots and complex numbers.

Introduction to Euler's Formula

Euler's formula, (e^{itheta} cos(theta) isin(theta)), is a fundamental tool in complex analysis. It allows us to express complex numbers in terms of exponentials, making it easier to manipulate and solve complex equations.

Problem Setup

Given the equation:

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

We will simplify and find the logarithm of this expression. Similarly, we will work on:

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

and then compare the logarithms of both expressions.

Simplifying the Expressions

Let's start by simplifying the first expression:

[frac{1 - sqrt{i}}{1 sqrt{i}} frac{1 - e^{frac{ipi}{4}}}{1 e^{frac{ipi}{4}}}]

Multiplying by the conjugate to rationalize the denominator:

[ frac{(1 - e^{frac{ipi}{4}})(1 - e^{-frac{ipi}{4}})}{(1 e^{frac{ipi}{4}})(1 - e^{-frac{ipi}{4}})} frac{1 - e^{frac{ipi}{4}} - e^{-frac{ipi}{4}} e^{frac{ipi}{4}} e^{-frac{ipi}{4}}}{2} frac{1 - 2cos(frac{pi}{4}) 1}{2} frac{2 - 2cdotfrac{sqrt{2}}{2}}{2} frac{2 - sqrt{2}}{2}]

Simplifying further:

[ frac{2 - sqrt{2}}{2} frac{2sqrt{2} - 2}{2sqrt{2}} frac{sqrt{2} - 1}{sqrt{2}}]

Thus, taking the logarithm:

[ilogleft(frac{1 - sqrt{i}}{1 sqrt{i}}right) ilogleft(frac{sqrt{2} - 1}{sqrt{2}}right) ileft(log(sqrt{2} - 1) - frac{pi}{2}iright) log(sqrt{2} - 1) frac{pi}{2}]

Second Expression

Now, let's handle the second expression:

[frac{1 - isqrt{i}}{1 isqrt{i}} frac{1 - e^{frac{3ipi}{4}}}{1 e^{frac{3ipi}{4}}}]

Again, rationalizing the denominator with the conjugate:

[ frac{(1 - e^{frac{3ipi}{4}})(1 - e^{-frac{3ipi}{4}})}{(1 e^{frac{3ipi}{4}})(1 - e^{-frac{3ipi}{4}})} frac{1 - e^{frac{3ipi}{4}} - e^{-frac{3ipi}{4}} e^{frac{3ipi}{4}} e^{-frac{3ipi}{4}}}{2} frac{1 - 2cos(frac{3pi}{4}) 1}{2} frac{2 - 2cdotfrac{sqrt{2}}{2}}{2} frac{2 - sqrt{2}}{2}]

Simplifying further:

[ frac{2 - sqrt{2}}{2} frac{2sqrt{2} - 2}{2sqrt{2}} frac{sqrt{2} - 1}{sqrt{2}}]

Thus, taking the logarithm:

[ilogleft(frac{1 - isqrt{i}}{1 isqrt{i}}right) ilogleft(frac{sqrt{2} - 1}{sqrt{2}}right) ileft(log(sqrt{2} - 1) - frac{pi}{2}iright) log(sqrt{2} - 1) frac{pi}{2}]

Comparing the Logarithms

Now, we compare the logarithms:

[ilogleft(frac{1 - sqrt{i}}{1 sqrt{i}}right) - ilogleft(frac{1 - isqrt{i}}{1 isqrt{i}}right) ileft(log(sqrt{2} - 1) frac{pi}{2}right) - ileft(log(sqrt{2} - 1) frac{pi}{2}right) -ileft(log(sqrt{2} - 1) - frac{pi}{2}right)]

This simplifies to:

[ -ileft(log(sqrt{2} - 1) - frac{pi}{2}right) ileft(frac{pi}{2} - log(sqrt{2} - 1)right)]

Final Calculation

Finally, we incorporate the final value:

[frac{1}{4isqrt{i}} frac{1}{4}e^{-frac{3pi}{4}} -frac{1 - i}{4sqrt{2}}]

Thus, the full calculation becomes:

[frac{1}{4isqrt{i}}left(ilogleft(frac{1 - sqrt{i}}{1 sqrt{i}}right) - ilogleft(frac{1 - isqrt{i}}{1 isqrt{i}}right)right) -frac{1 - i}{4sqrt{2}} cdot ileft(frac{pi}{2} - log(sqrt{2} - 1)right) -frac{1 - i}{4sqrt{2}} cdot ileft(frac{pi}{2} - frac{pi}{2}right) -frac{1 - i}{4sqrt{2}} cdot 2left(frac{pi}{2} - log(sqrt{2} - 1)right)]

This further simplifies to:

[ -frac{1 - i}{4sqrt{2}} cdot 2left(frac{pi}{2} - log(sqrt{2} - 1)right) -frac{1 - i}{4sqrt{2}} cdot 2left(frac{pi}{2} - log(1 - frac{1}{sqrt{2}})right) -frac{1 - i}{4sqrt{2}} cdot 2left(frac{pi}{2} log(frac{1}{sqrt{2} - 1})right)]

Finally:

[ -frac{1 - i}{4sqrt{2}} cdot 2left(frac{pi}{2} - log(frac{1}{sqrt{2} - 1})right) -frac{pi}{4sqrt{2}} - frac{pi}{2sqrt{2}} frac{pi - 2log(sqrt{2} - 1)}{4sqrt{2}} -frac{pi - 2log(frac{3}{2sqrt{2}})}{4sqrt{2}}]

Thus, the final boxed result is:

[boxed{-frac{pilog(3) - 2sqrt{2}}{4sqrt{2}}}]