Convert the following in the polar form:
`(1+7i)/((2-i)^(2))`

Answer

To convert the expression \((1 + 7i) / ((2 - i)^2)\) into polar form, we will follow these steps:

### Step 1: Simplify the denominator
First, we need to calculate \((2 - i)^2\).

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

### Step 2: Rewrite the expression
Now we can rewrite the original expression:

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

### Step 3: Multiply by the conjugate of the denominator
To simplify the fraction, we multiply the numerator and the denominator by the conjugate of the denominator, which is \(3 + 4i\):

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

### Step 4: Simplify the denominator
Calculating the denominator:

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

### Step 5: Simplify the numerator
Now, we calculate the numerator:

\[ (1 + 7i)(3 + 4i) = 1 \cdot 3 + 1 \cdot 4i + 7i \cdot 3 + 7i \cdot 4i = 3 + 4i + 21i + 28i^2 \]
\[ = 3 + 25i + 28(-1) = 3 + 25i - 28 = -25 + 25i \]

### Step 6: Combine the results
Now we can combine the results:

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

### Step 7: Convert to polar form
To convert \(-1 + i\) into polar form, we need to find the modulus \(r\) and the argument \(\theta\).

#### Finding the modulus \(r\):

\[ r = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

#### Finding the argument \(\theta\):
The argument \(\theta\) can be found using:

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

This indicates that the angle is in the second quadrant. The reference angle where \(\tan(\theta) = 1\) is \(\frac{\pi}{4}\), thus:

\[ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]

### Final Polar Form
Now we can write the polar form:

\[ \text{Polar form} = r(\cos \theta + i \sin \theta) = \sqrt{2} \left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right) \]

### Summary
The polar form of \(\frac{1 + 7i}{(2 - i)^2}\) is:

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

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