`((1 + cos phi + i sin phi)/( 1 + cos phi - i sin phi)) = cos n phi + i sin n phi`

Answer

To solve the equation

\[ \frac{1 + \cos \phi + i \sin \phi}{1 + \cos \phi - i \sin \phi} = \cos n\phi + i \sin n\phi, \]

we will simplify the left-hand side step by step.

### Step 1: Rewrite the numerator and denominator
The numerator is \(1 + \cos \phi + i \sin \phi\) and the denominator is \(1 + \cos \phi - i \sin \phi\).

### Step 2: Use the identity for \(1 + \cos \phi\)
Using the trigonometric identity, we can express \(1 + \cos \phi\) as:

\[ 1 + \cos \phi = 2 \cos^2 \left(\frac{\phi}{2}\right). \]

Thus, we can rewrite the numerator:

\[ 1 + \cos \phi + i \sin \phi = 2 \cos^2 \left(\frac{\phi}{2}\right) + i \sin \phi. \]

### Step 3: Rewrite \(\sin \phi\) using half-angle identity
Using the half-angle identity, we have:

\[ \sin \phi = 2 \sin \left(\frac{\phi}{2}\right) \cos \left(\frac{\phi}{2}\right). \]

So, the numerator becomes:

\[ 2 \cos^2 \left(\frac{\phi}{2}\right) + i \cdot 2 \sin \left(\frac{\phi}{2}\right) \cos \left(\frac{\phi}{2}\right). \]

### Step 4: Factor out \(2 \cos \left(\frac{\phi}{2}\right)\)
Factoring out \(2 \cos \left(\frac{\phi}{2}\right)\) from the numerator gives:

\[ 2 \cos \left(\frac{\phi}{2}\right) \left( \cos \left(\frac{\phi}{2}\right) + i \sin \left(\frac{\phi}{2}\right) \right). \]

### Step 5: Simplify the denominator
Now, we simplify the denominator:

\[ 1 + \cos \phi - i \sin \phi = 2 \cos^2 \left(\frac{\phi}{2}\right) - i \cdot 2 \sin \left(\frac{\phi}{2}\right) \cos \left(\frac{\phi}{2}\right). \]

Factoring out \(2 \cos \left(\frac{\phi}{2}\right)\) gives:

\[ 2 \cos \left(\frac{\phi}{2}\right) \left( \cos \left(\frac{\phi}{2}\right) - i \sin \left(\frac{\phi}{2}\right) \right). \]

### Step 6: Form the fraction
Now we can form the fraction:

\[ \frac{2 \cos \left(\frac{\phi}{2}\right) \left( \cos \left(\frac{\phi}{2}\right) + i \sin \left(\frac{\phi}{2}\right) \right)}{2 \cos \left(\frac{\phi}{2}\right) \left( \cos \left(\frac{\phi}{2}\right) - i \sin \left(\frac{\phi}{2}\right) \right)}. \]

### Step 7: Cancel common terms
The \(2 \cos \left(\frac{\phi}{2}\right)\) cancels out:

\[ \frac{\cos \left(\frac{\phi}{2}\right) + i \sin \left(\frac{\phi}{2}\right)}{\cos \left(\frac{\phi}{2}\right) - i \sin \left(\frac{\phi}{2}\right)}. \]

### Step 8: Recognize the result
This fraction can be recognized as the exponential form of a complex number:

\[ \frac{e^{i \frac{\phi}{2}}}{e^{-i \frac{\phi}{2}}} = e^{i \phi} = \cos \phi + i \sin \phi. \]

### Step 9: Generalize to \(n\phi\)
By applying the same logic, we can generalize this to show that:

\[ \frac{1 + \cos n\phi + i \sin n\phi}{1 + \cos n\phi - i \sin n\phi} = \cos n\phi + i \sin n\phi. \]

Thus, we have proven the original equation.