Find the modulus and amplitude of the following complex numbers and hence express them into polar form
`((1+i)^(13))/((1-i)^(7))`

Answer

To find the modulus and amplitude of the complex number \( z = \frac{(1+i)^{13}}{(1-i)^{7}} \) and express it in polar form, we will follow these steps:

### Step 1: Calculate the modulus of \( 1+i \) and \( 1-i \)

The modulus of a complex number \( a + bi \) is given by \( |z| = \sqrt{a^2 + b^2} \).

For \( 1+i \):
\[ |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

For \( 1-i \):
\[ |1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]

### Step 2: Calculate the modulus of \( z \)

Using the property of moduli:
\[ |z| = \frac{|(1+i)^{13}|}{|(1-i)^{7}|} = \frac{|1+i|^{13}}{|1-i|^{7}} = \frac{(\sqrt{2})^{13}}{(\sqrt{2})^{7}} = \frac{2^{13/2}}{2^{7/2}} = 2^{(13/2 - 7/2)} = 2^{6/2} = 2^3 = 8 \]

### Step 3: Calculate the amplitude (argument) of \( 1+i \) and \( 1-i \)

The argument of a complex number \( a + bi \) is given by \( \tan^{-1}(\frac{b}{a}) \).

For \( 1+i \):
\[ \text{arg}(1+i) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]

For \( 1-i \):
\[ \text{arg}(1-i) = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \]

### Step 4: Calculate the argument of \( z \)

Using the property of arguments:
\[ \text{arg}(z) = \text{arg}((1+i)^{13}) - \text{arg}((1-i)^{7}) = 13 \cdot \text{arg}(1+i) - 7 \cdot \text{arg}(1-i) \]
\[ = 13 \cdot \frac{\pi}{4} - 7 \cdot \left(-\frac{\pi}{4}\right) = \frac{13\pi}{4} + \frac{7\pi}{4} = \frac{20\pi}{4} = 5\pi \]

### Step 5: Express \( z \) in polar form

The polar form of a complex number is given by:
\[ z = r(\cos \theta + i \sin \theta) \]
Where \( r \) is the modulus and \( \theta \) is the argument.

Thus, we have:
\[ z = 8 \left( \cos(5\pi) + i \sin(5\pi) \right) \]

Since \( \cos(5\pi) = -1 \) and \( \sin(5\pi) = 0 \), we can simplify:
\[ z = 8(-1 + 0i) = -8 \]

### Final Result

The modulus of \( z \) is \( 8 \), the argument is \( 5\pi \), and in polar form, \( z = 8(\cos(5\pi) + i \sin(5\pi)) \).

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