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Mobile App Chapter 2: Problem 19
Verify the formulas. $$\arctan z=\frac{1}{2 i} \ln \frac{1+i z}{1-i z}$$
Short Answer
Expert verified
The formula is correct.
Step by step solution
01
- Understand the formula
We need to verify the given formula for \(\text{arctan} \ z \): \[ \arctan(z) = \frac{1}{2i} \ln \frac{1 + i z}{1 - i z}\].
02
- Define \(z = \tan(\theta)\)
To verify the formula, assume \(\theta\) such that \( z = \ \tan(\theta) \). Therefore, \( \theta = \text{arctan}(z)\).
03
- Apply Identities
Utilize trigonometric identities to rewrite the formula. Specifically, \( \tan(\theta) = z \Rightarrow \1 + i \tan(\theta) = \frac{\cos(\theta) + i \sin(\theta)}{\cos(\theta)}\) and \(1 - i \tan(\theta) = \frac{\cos(\theta) - i \sin(\theta)}{\cos(\theta)}\).
04
- Simplify the Logarithm Argument
Using the expressions from Step 3, express the argument of the logarithm: \[ \frac{1+i z}{1-i z} = \frac{ \cos(\theta) + i \sin(\theta) }{\cos(\theta) - i \sin(\theta)}.\]
05
- Use the Exponential Form
Recall that in exponential form, \(\theta\) can be expressed as \(\tan(\theta) = \frac{\text{Im}(e^{i\theta})}{\text{Re}(e^{i\theta})} \). Hence the argument converts as follows: \[ \frac{ \cos(\theta) + i \sin(\theta)}{\cos(\theta) - i \sin(\theta)} = e^{i \theta}.\]
06
- Apply Logarithm Property
Using properties of complex logarithms, rewrite the expression: \[ \ln \left(e^{i \theta} \right) = i \theta.\]
07
- Put it All Together
Substitute \( i \theta\) into the original formula: \[ \arctan(z) = \theta = \frac{1}{2i} \ln \frac{1 + i z}{1 - i z}\].\ Thus, the formula is verified.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
arctan formula
The \arctan\ formula presented is: \( \text{arctan}(z) = \frac{1}{2i} \text{ln} \frac{1 + i z}{1 - i z}\). This might look intricate at first glance, but let's break it down step by step for better understanding. The arctan or inverse tangent function helps determine the angle whose tangent value is z. By transforming the arctan formula into a logarithmic form, we are exploring its relationships in the realm of complex numbers. To verify this, we employ trigonometric identities and properties of logarithms. This aids in simplifying and connecting complex numbers to trigonometric functions.
trigonometric identities
Trigonometric identities play a crucial role in transforming and simplifying the expression. Key identities used include:
- \( \tan(\theta) = \frac{\text{sin}(\theta)}{\text{cos}(\theta)} \)
- \( e^{i \theta} = \text{cos}(\theta) + i \text{sin}(\theta) \)
exponential form
Expressing trigonometric functions in exponential form often simplifies complex expressions. Using Euler's formula, \( e^{i \theta} = \text{cos}(\theta) + i \text{sin}(\theta) \), allows us to handle complex logarithms more easily. Here’s how:
- \( \theta = \text{arctan}(z) \) implies that \( e^{i \theta} \) represents a point on the unit circle.
- Consequently, \( \text{ln}(e^{i \theta}) = i \theta \) by properties of logarithms.
