Convert the complex number `(-16)/(1+isqrt(3))`into polar form.

Explanation

To convert the complex number \(\frac{-16}{1 + i\sqrt{3}}\) into polar form, we will follow these steps:

### Step 1: Rationalize the denominator
We start by multiplying the numerator and the denominator by the conjugate of the denominator:

\[ \frac{-16}{1 + i\sqrt{3}} \cdot \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}} = \frac{-16(1 - i\sqrt{3})}{(1 + i\sqrt{3})(1 - i\sqrt{3})} \]

### Step 2: Simplify the denominator
Now, we simplify the denominator using the difference of squares formula:

\[ (1 + i\sqrt{3})(1 - i\sqrt{3}) = 1^2 - (i\sqrt{3})^2 = 1 - (-3) = 1 + 3 = 4 \]

### Step 3: Simplify the numerator
Next, we simplify the numerator:

\[ -16(1 - i\sqrt{3}) = -16 + 16i\sqrt{3} \]

### Step 4: Combine the results
Now we can write the expression as:

\[ \frac{-16 + 16i\sqrt{3}}{4} = -4 + 4i\sqrt{3} \]

### Step 5: Write in standard form
Now we have the complex number in standard form:

\[ -4 + 4i\sqrt{3} \]

### Step 6: Find the modulus
To convert to polar form, we need to find the modulus \(r\):

\[ r = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \]

### Step 7: Find the argument
Next, we find the argument \(\theta\):

\[ \tan(\theta) = \frac{\text{Imaginary part}}{\text{Real part}} = \frac{4\sqrt{3}}{-4} = -\sqrt{3} \]

This corresponds to an angle in the second quadrant. The reference angle where \(\tan(\theta) = \sqrt{3}\) is \(\frac{\pi}{3}\). Thus, the angle in the second quadrant is:

\[ \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

### Step 8: Write in polar form
Finally, we can express the complex number in polar form:

\[ 8 \left( \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) \right) \]

### Final Answer
The polar form of the complex number \(\frac{-16}{1 + i\sqrt{3}}\) is:

\[ 8 \text{cis} \left(\frac{2\pi}{3}\right) \]

where \(\text{cis}(\theta) = \cos(\theta) + i\sin(\theta)\).

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