Convert the following in polar form,
(iv)`(5-i)/(2-3i)`

Answer

To convert the complex number \(\frac{5 - i}{2 - 3i}\) into polar form, we will follow these steps:

### Step 1: Rationalize the denominator
We start with the expression:
\[ z = \frac{5 - i}{2 - 3i} \]
To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is \(2 + 3i\):
\[ z = \frac{(5 - i)(2 + 3i)}{(2 - 3i)(2 + 3i)} \]

### Step 2: Simplify the denominator
Using the formula for the difference of squares, \(a^2 - b^2\):
\[ (2 - 3i)(2 + 3i) = 2^2 - (3i)^2 = 4 - 9(-1) = 4 + 9 = 13 \]

### Step 3: Expand the numerator
Now, we expand the numerator:
\[ (5 - i)(2 + 3i) = 5 \cdot 2 + 5 \cdot 3i - i \cdot 2 - i \cdot 3i = 10 + 15i - 2i - 3(-1) \]
This simplifies to:
\[ 10 + 15i - 2i + 3 = 13 + 13i \]

### Step 4: Combine the results
Now we can write \(z\) as:
\[ z = \frac{13 + 13i}{13} = 1 + i \]

### Step 5: Convert to polar form
To convert \(z = 1 + i\) into polar form, we need to find the modulus \(r\) and the argument \(\theta\).

#### Finding the modulus \(r\):
\[ r = |z| = \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

#### Finding the argument \(\theta\):
The argument \(\theta\) can be found using:
\[ \tan(\theta) = \frac{b}{a} = \frac{1}{1} = 1 \]
Thus, \(\theta = \frac{\pi}{4}\) (since both \(a\) and \(b\) are positive, we are in the first quadrant).

### Final Polar Form
Now we can express \(z\) in polar form:
\[ z = r(\cos \theta + i \sin \theta) = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \]

### Summary
The polar form of \(\frac{5 - i}{2 - 3i}\) is:
\[ \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \]

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