Represent the complex numbers
`(1+7i)/((2-i)^(2))` in polar form

Answer

To represent the complex number \(\frac{1 + 7i}{(2 - i)^2}\) in polar form, we will follow these steps:

### Step 1: Calculate \((2 - i)^2\)
We start by squaring the denominator:

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

### Step 2: Rewrite the complex number
Now we can rewrite the complex number:

\[ z = \frac{1 + 7i}{3 - 4i} \]

### Step 3: Rationalize the denominator
To simplify the expression, we multiply the numerator and denominator by the conjugate of the denominator:

\[ z = \frac{(1 + 7i)(3 + 4i)}{(3 - 4i)(3 + 4i)} \]

Calculating the denominator:

\[ (3 - 4i)(3 + 4i) = 3^2 - (4i)^2 = 9 - 16(-1) = 9 + 16 = 25 \]

Now for the numerator:

\[ (1 + 7i)(3 + 4i) = 1 \cdot 3 + 1 \cdot 4i + 7i \cdot 3 + 7i \cdot 4i = 3 + 4i + 21i + 28(-1) = 3 + 25i - 28 = -25 + 25i \]

### Step 4: Combine the results
Putting it all together, we have:

\[ z = \frac{-25 + 25i}{25} = -1 + i \]

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

1. **Modulus**:
\[ r = |z| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

2. **Argument**:
\[ \theta = \tan^{-1}\left(\frac{1}{-1}\right) = \tan^{-1}(-1) \]
Since the complex number \(-1 + i\) lies in the second quadrant, we adjust the angle:
\[ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]

### Final Result
Thus, the polar form of the complex number is:

\[ z = \sqrt{2} \left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right) \]