Chapter 2: Problem 10
Evaluate each of the following in \(x+i y\) form, and compare with a computer solution. $$i^{\ln i}$$
Short Answer
Expert verified
The expression evaluates to \(e^{-\frac{\pi}{2}}\).
Step by step solution
01
Understand the Problem
The task is to evaluate the expression \( i^{\ln i} \) and express it in the form \( x + i y \), where \(x \) and \( y \) are real numbers.
02
Apply the Exponential and Logarithm Rules
Recall that \(i = e^{i \frac{\pi}{2}} \). Using this, \(\ln i\) can be found as follows: \({\ln i = i \frac{\pi}{2}} \).
03
Substitute the Logarithmic Value
Now, substitute the value of \(\ln i\) into the given expression: \( i^{\ln i} = i^{i \frac{\pi}{2}} \).
04
Use the Euler's Formula
Using Euler's formula \(e^{i \theta} = \cos \theta + i \sin \theta\), the expression \(i^{i \frac{\pi}{2}} \) can be written as \(e^{i (i \frac{\pi}{2})} = e^{-\frac{\pi}{2}} \).
05
Simplify the Expression
Since \(e^{-\frac{\pi}{2}}\) is a real number, the final simplified form is: \(e^{-\frac{\pi}{2}} + 0i\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
complex exponents
Complex exponents can be tricky, but they follow rules similar to real exponents. Let's break it down with an example. Suppose we have an expression like \(i^{i}\). To understand this, we use the fact that any complex number can be expressed in the exponential form. For instance, \(i\) can be written as \(e^{i\frac{\pi}{2}}\) because Euler's formula tells us that \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\). This is crucial because it allows us to convert between different forms more easily.
Next, consider the logarithm of a complex number. Specifically, for our problem, we find that \( \ln i = i \frac{\pi}{2}\). Why? Because taking the natural logarithm is the inverse operation of the exponential. We essentially find the angle (or argument) of the complex number in the complex plane. Using these transformations helps us simplify otherwise complicated expressions.
Substituting these values back into our initial expression, we have: \(i^{\ln i} = i^{i \frac{\pi}{2}}\). This might still look complicated, but if you use the rewritten form of \(i\) and apply the exponential rules, it becomes much simpler.
Next, consider the logarithm of a complex number. Specifically, for our problem, we find that \( \ln i = i \frac{\pi}{2}\). Why? Because taking the natural logarithm is the inverse operation of the exponential. We essentially find the angle (or argument) of the complex number in the complex plane. Using these transformations helps us simplify otherwise complicated expressions.
Substituting these values back into our initial expression, we have: \(i^{\ln i} = i^{i \frac{\pi}{2}}\). This might still look complicated, but if you use the rewritten form of \(i\) and apply the exponential rules, it becomes much simpler.
Euler's formula
Euler's formula is a fundamental bridge between exponential functions and trigonometry. It’s expressed as \(e^{i \theta} = \cos \theta + i \sin \theta\). This is not just a formula but a gateway to understanding complex numbers deeply.
For example, when dealing with \(i = e^{i\frac{\pi}{2}}\), we are using Euler's formula. The angle \(\frac{\pi}{2}\) is crucial because when plugged into Euler’s formula, it results in \(\cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2})\), where \(\cos(\frac{\pi}{2}) = 0\) and \(\sin(\frac{\pi}{2}) = 1\), giving us \(0 + i\), which is \(i\).
This formula allows us to switch between exponential and trigonometric forms of complex numbers seamlessly. In our problem, we use Euler's formula to handle the exponentiation of a complex number. After we write \(i\) as \(e^{i\frac{\pi}{2}}\), applying the formula leads to easy simplification. Through this, we transformed \(i^{i \frac{\pi}{2}}\) into \(e^{i (i \frac{\pi}{2})}\). This quickly simplifies to \(e^{ - \frac{\pi}{2}}\).
For example, when dealing with \(i = e^{i\frac{\pi}{2}}\), we are using Euler's formula. The angle \(\frac{\pi}{2}\) is crucial because when plugged into Euler’s formula, it results in \(\cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2})\), where \(\cos(\frac{\pi}{2}) = 0\) and \(\sin(\frac{\pi}{2}) = 1\), giving us \(0 + i\), which is \(i\).
This formula allows us to switch between exponential and trigonometric forms of complex numbers seamlessly. In our problem, we use Euler's formula to handle the exponentiation of a complex number. After we write \(i\) as \(e^{i\frac{\pi}{2}}\), applying the formula leads to easy simplification. Through this, we transformed \(i^{i \frac{\pi}{2}}\) into \(e^{i (i \frac{\pi}{2})}\). This quickly simplifies to \(e^{ - \frac{\pi}{2}}\).
natural logarithm
The natural logarithm is typically thought of with real numbers, but it also applies to complex numbers. For real numbers, the natural logarithm function reverses the exponential function: \(\ln(e^x) = x\). With complex numbers, it’s similar but requires consideration of the complex plane.
For complex number \(i\), its natural logarithm is \( \ln(i) = i \frac{\pi}{2}\). This arises because logarithms convert multiplication into addition of their arguments. Remember that \(i\) can be written as \(e^{i \frac{\pi}{2}}\), so \(\ln(i) = \ln(e^{i \frac{\pi}{2}})\), which simplifies directly to \(i \frac{\pi}{2}\).
In our given problem, evaluating \(i^{\ln(i)}\) involves first finding \(\ln(i)\), which we determined as \(i \frac{\pi}{2}\). Then, we substitute it back into the exponent: \(i^{i \frac{\pi}{2}}\). Using our knowledge of exponent rules and simplifying with Euler's formula, we get \(e^{ - \frac{\pi}{2}}\), showing the natural logarithm’s crucial role in understanding complex exponents.
For complex number \(i\), its natural logarithm is \( \ln(i) = i \frac{\pi}{2}\). This arises because logarithms convert multiplication into addition of their arguments. Remember that \(i\) can be written as \(e^{i \frac{\pi}{2}}\), so \(\ln(i) = \ln(e^{i \frac{\pi}{2}})\), which simplifies directly to \(i \frac{\pi}{2}\).
In our given problem, evaluating \(i^{\ln(i)}\) involves first finding \(\ln(i)\), which we determined as \(i \frac{\pi}{2}\). Then, we substitute it back into the exponent: \(i^{i \frac{\pi}{2}}\). Using our knowledge of exponent rules and simplifying with Euler's formula, we get \(e^{ - \frac{\pi}{2}}\), showing the natural logarithm’s crucial role in understanding complex exponents.
