Find the modulus and amplitude of the following complex numbers and hence express them into polar form
`(5-i)/(2-3i)`

Answer

To find the modulus and amplitude of the complex number \( z = \frac{5 - i}{2 - 3i} \) and express it in polar form, we will follow these steps:

### Step 1: Multiply by the Conjugate
To simplify the division of complex numbers, we multiply the numerator and denominator by the conjugate of the denominator. The conjugate of \( 2 - 3i \) is \( 2 + 3i \).

\[ z = \frac{5 - i}{2 - 3i} \cdot \frac{2 + 3i}{2 + 3i} \]

### Step 2: Simplify the Numerator
Now, we calculate the numerator:

\[ (5 - i)(2 + 3i) = 5 \cdot 2 + 5 \cdot 3i - i \cdot 2 - i \cdot 3i = 10 + 15i - 2i - 3i^2 \]

Since \( i^2 = -1 \), we have:

\[ -3i^2 = 3 \]

Thus, the numerator simplifies to:

\[ 10 + 15i - 2i + 3 = 13 + 13i \]

### Step 3: Simplify the Denominator
Now, we calculate the denominator:

\[ (2 - 3i)(2 + 3i) = 2^2 - (3i)^2 = 4 - 9(-1) = 4 + 9 = 13 \]

### Step 4: Combine the Results
Now we can combine the results from the numerator and denominator:

\[ z = \frac{13 + 13i}{13} = 1 + i \]

### Step 5: Find the Modulus
The modulus of a complex number \( a + bi \) is given by:

\[ |z| = \sqrt{a^2 + b^2} \]

For \( z = 1 + i \):

\[ |z| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 6: Find the Amplitude (Argument)
The amplitude (or argument) \( \theta \) is given by:

\[ \tan \theta = \frac{\text{Imaginary part}}{\text{Real part}} = \frac{1}{1} = 1 \]

Thus,

\[ \theta = \tan^{-1}(1) = \frac{\pi}{4} \]

### Step 7: Express 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. Therefore, we have:

\[ z = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \]

### Final Result
The modulus of \( z \) is \( \sqrt{2} \), the amplitude is \( \frac{\pi}{4} \), and the polar form is:

\[ z = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \]

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