Change the following complex numbers into polar form
`(1+ 2i)/(1-(1-i)^(2))`

Answer

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

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

Using the formula for squaring a binomial, we have:
\[ (1 - i)^2 = 1^2 - 2 \cdot 1 \cdot i + i^2 = 1 - 2i + (-1) = 0 - 2i = -2i \]
Thus, the denominator becomes:
\[ 1 - (1 - i)^2 = 1 - (-2i) = 1 + 2i \]

### Step 2: Rewrite the complex number
Now we can rewrite the complex number \(z\):
\[ z = \frac{1 + 2i}{1 + 2i} \]

### Step 3: Simplify the complex number
Since the numerator and the denominator are the same, we can simplify:
\[ z = 1 \]

### Step 4: Express in polar form
The polar form of a complex number \(z\) is given by:
\[ z = r(\cos \theta + i \sin \theta) \]
where \(r\) is the modulus and \(\theta\) is the argument.

### Step 5: Calculate the modulus
The modulus of \(z = 1\) is:
\[ r = |z| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 \]

### Step 6: Calculate the argument
The argument \(\theta\) is given by:
\[ \theta = \tan^{-1}\left(\frac{0}{1}\right) = \tan^{-1}(0) = 0 \]

### Step 7: Write the polar form
Now we can express \(z\) in polar form:
\[ z = 1(\cos 0 + i \sin 0) \]
This can be simplified to:
\[ z = \cos 0 + i \sin 0 \]

### Final Answer
Thus, the polar form of the given complex number is:
\[ z = 1 \text{ (or } \cos 0 + i \sin 0\text{)} \]

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