Evaluating a Complex Number Expression when \(z = e^{i\theta}\)

This problem involves simplifying an expression containing a complex number \(z\). The complex number is given in exponential form, \(z = e^{i\theta}\). We need to find the value of \(\frac{{{z^2} - 1}}{{{z^2} + 1}}\).

Understanding the Given Complex Number

We are given the complex number \(z\) in exponential form:

\(z = e^{i\theta}\)

To work with this, we can use Euler's formula, which relates the exponential form of a complex number to its trigonometric form:

\(e^{ix} = \cos x + i\sin x\)

Applying Euler's formula to \(z\), we get:

\(z = \cos\theta + i\sin\theta\)

Calculating \(z^2\) using Euler's Formula

First, let's calculate \(z^2\). Using the exponential form is often easier for powers:

\(z^2 = (e^{i\theta})^2 = e^{i(2\theta)}\)

Now, applying Euler's formula again to \(e^{i(2\theta)}\):

\(z^2 = \cos(2\theta) + i\sin(2\theta)\)

This gives us \(z^2\) in trigonometric form, which is useful for the next steps.

Substituting \(z^2\) into the Expression

Now we substitute the expression for \(z^2\) into the given fraction \(\frac{{{z^2} - 1}}{{{z^2} + 1}}\):

\(\frac{{{z^2} - 1}}{{{z^2} + 1}} = \frac{(\cos(2\theta) + i\sin(2\theta)) - 1}{(\cos(2\theta) + i\sin(2\theta)) + 1}\)

Rearrange the terms in the numerator and denominator to group real and imaginary parts:

\(\frac{(\cos(2\theta) - 1) + i\sin(2\theta)}{(\cos(2\theta) + 1) + i\sin(2\theta)}\)

Applying Trigonometric Identities for Simplification

To simplify further, we use standard double-angle trigonometric identities:

• \(\cos(2\theta) - 1 = -2\sin^2\theta\)
• \(\cos(2\theta) + 1 = 2\cos^2\theta\)
• \(\sin(2\theta) = 2\sin\theta\cos\theta\)

Substitute these identities into the expression:

\(\frac{(-2\sin^2\theta) + i(2\sin\theta\cos\theta)}{(2\cos^2\theta) + i(2\sin\theta\cos\theta)}\)

Factoring and Canceling Terms

Now, we can factor out common terms from the numerator and the denominator. From the numerator, we can factor out \(2\sin\theta\):

Numerator: \(-2\sin^2\theta + i(2\sin\theta\cos\theta) = 2\sin\theta(-\sin\theta + i\cos\theta)\)

From the denominator, we can factor out \(2\cos\theta\):

Denominator: \(2\cos^2\theta + i(2\sin\theta\cos\theta) = 2\cos\theta(\cos\theta + i\sin\theta)\)

Substitute these factored forms back into the fraction:

\(\frac{2\sin\theta(-\sin\theta + i\cos\theta)}{2\cos\theta(\cos\theta + i\sin\theta)}\)

Cancel the common factor of 2:

\(\frac{\sin\theta(-\sin\theta + i\cos\theta)}{\cos\theta(\cos\theta + i\sin\theta)}\)

Simplifying the Complex Number Term

Look at the term \((-\sin\theta + i\cos\theta)\) in the numerator. We can factor out \(i\):

\(-\sin\theta + i\cos\theta = i\left(\frac{-\sin\theta}{i} + \cos\theta\right)\)

Since \(\frac{1}{i} = -i\), we have \(\frac{-\sin\theta}{i} = -\sin\theta (-i) = i\sin\theta\).

So, \((-\sin\theta + i\cos\theta) = i(i\sin\theta + \cos\theta) = i(\cos\theta + i\sin\theta)\).

Now, substitute this back into the fraction:

\(\frac{\sin\theta \cdot i(\cos\theta + i\sin\theta)}{\cos\theta(\cos\theta + i\sin\theta)}\)

We can now cancel the common complex number term \((\cos\theta + i\sin\theta)\) from both the numerator and the denominator (assuming \(\cos\theta + i\sin\theta \ne 0\), which is true for any real \(\theta\)):

\(i\frac{\sin\theta}{\cos\theta}\)

Final Result for the Complex Number Expression

Using the identity \(\frac{\sin\theta}{\cos\theta} = \tan\theta\), the expression simplifies to:

\(i\tan\theta\)

This is the final value of the given complex number expression.

Summary of Steps for this Complex Number Problem

To solve this complex number problem:

1. Recognize the complex number form \(z = e^{i\theta}\).
2. Use Euler's formula to find \(z^2\).
3. Substitute \(z^2\) into the expression.
4. Apply trigonometric identities for \(\cos(2\theta)\) and \(\sin(2\theta)\).
5. Factor out common terms.
6. Simplify the remaining complex number terms and trigonometric ratio.

The result is \(i\tan\theta\).