Question

Find one or more values of each of the following complex expressions in the easiest way you can. \(\cos \left[2 i \ln \frac{1-i}{1+i}\right]\)

Short answer

shortanswer: simplify.

Step-by-step solution

- Simplify the Argument Inside the Cosine Function

We use properties of logarithms and exponentials to further simplify:

- Evaluate the Simplified Argument

Now, we have the expression: t\[\t\oi \left[2i \times \ln (i(1i))\right]\].Next, we evaluate the natural logarithm:\tt\[\oi (i)=\log (\oe )=\phi (1+i=\frac{LP (\frac(0,1)}{(i+))0power of).\right)\].in the cosine argument simplicity propertiescontinue: step 2 next - simplify logarithmfinish

Key concepts

Complex Expressions

Complex expressions are made up of complex numbers, containing both real and imaginary parts. For a number in the form of \( a + bi \), \(a\) is the real part, and \(bi\) is the imaginary part.
When working with complex expressions, it's essential to understand the operations involved:

• Addition: Combine like terms by adding the real parts and the imaginary parts separately.
• Multiplication: Use the distributive property and remember that \(i^2 = -1\).
• Division: Multiply the numerator and the denominator by the conjugate of the denominator to remove the imaginary part from the denominator.

This exercise involves using the cosine function on a complex argument. Simplifying complex expressions before applying trigonometric functions is crucial.

Logarithms

Logarithms help to simplify complex expressions by transforming multiplicative relationships into additive ones. The natural logarithm, denoted \( \ln \, \), is particularly useful.

• The natural logarithm of a product can be split into the sum of the logarithms: \( \ln(ab) = \ln(a) + \ln(b) \).
• For division, the logarithm of a quotient is the difference between the logarithms: \( \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b) \).
• When raising a number to a power, the logarithm becomes a product: \( \ln(a^b) = b \ln(a) \).

In the given problem, \( \ln \left( \frac{1 - i}{1 + i} \right) \) is simplified using these properties. Consider that \(1 - i\) and \(1 + i\) can be evaluated using their magnitudes and angles. This step helps in eventually evaluating the argument inside the cosine function.

Exponential Functions

Exponential functions denote a constant raised to the power of a variable. They are central to linking logarithms and trigonometric functions:

• The function \( e^{ix} = \cos(x) + i \sin(x) \) forms the basis of Euler's formula, crucial for simplifying complex exponentials.
• The inverse, \( \ln(e^x) = x \), shows how logarithms and exponentials undo each other.
• Exponential functions simplify complex multiplication and division.

In this exercise, recognizing the exponential form is vital. Simplifying \( e^{i \theta} \) as \( \cos(\theta) + i \sin(\theta) \) assists in transforming complex expressions into trigonometric functions.

Cosine Function

Cosine function comes from trigonometry and is used to represent the horizontal component of a circular motion. When dealing with complex arguments, the cosine function ties into Euler's formula:

• For a real number \(x\), \( \cos(x) \) gives the horizontal position on the unit circle.
• For a complex argument, \( \cos(2i \ln(y)) \) requires transforming the argument into a real number.
• Euler's formula \( e^{ix} = \cos(x) + i \sin(x) \) simplifies transformations involving cosine.

In the given problem, understanding \( \cos \left[2 i \ln \frac{1-i}{1+i} \right] \) requires breaking down the \( \ln \) part, simplifying the expression, and then applying the cosine function to the simplified real argument.