Convert the complex number `4/(1-isqrt3)` in the polar form.

Answer

To convert the complex number \( \frac{4}{1 - i\sqrt{3}} \) into polar form, we will follow these steps:

### Step 1: Rationalize the Complex Number
We will multiply the numerator and the denominator by the conjugate of the denominator, which is \( 1 + i\sqrt{3} \).

\[ \frac{4}{1 - i\sqrt{3}} \cdot \frac{1 + i\sqrt{3}}{1 + i\sqrt{3}} = \frac{4(1 + i\sqrt{3})}{(1 - i\sqrt{3})(1 + i\sqrt{3})} \]

### Step 2: Simplify the Denominator
Using the difference of squares formula, we can simplify the denominator:

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

### Step 3: Simplify the Numerator
Now, we simplify the numerator:

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

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

\[ \frac{4 + 4i\sqrt{3}}{4} = 1 + i\sqrt{3} \]

### Step 5: Find the Modulus \( r \)
The modulus \( r \) of the complex number \( 1 + i\sqrt{3} \) is calculated as follows:

\[ r = |z| = \sqrt{x^2 + y^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 6: Find the Argument \( \theta \)
The argument \( \theta \) can be found using:

\[ \tan(\theta) = \frac{y}{x} = \frac{\sqrt{3}}{1} \]

Thus,

\[ \theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \text{ radians (or } 60^\circ\text{)} \]

### Step 7: Write in Polar Form
The polar form of a complex number is given by:

\[ r(\cos \theta + i \sin \theta) \]

Substituting the values of \( r \) and \( \theta \):

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

### Final Result
Thus, the polar form of the complex number \( \frac{4}{1 - i\sqrt{3}} \) is:

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

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