The angle that the vector representing the complex number `1/(sqrt3-i)^25` makes with the positive direction of the real axis is :

Options

A. `(2pi)/3`
B. `pi/6`
C. `(4pi)/9`
D. `(5pi)/6`

Answer

To find the angle that the vector representing the complex number \( \frac{1}{(\sqrt{3} - i)^{25}} \) makes with the positive direction of the real axis, we can follow these steps:

### Step 1: Simplify the Denominator
First, we need to express \( \sqrt{3} - i \) in polar form. We can find the modulus and argument of the complex number.

The modulus \( r \) of \( \sqrt{3} - i \) is given by:
\[ r = | \sqrt{3} - i | = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2. \]

### Step 2: Find the Argument
Next, we find the argument \( \theta \):
\[ \theta = \tan^{-1} \left( \frac{-1}{\sqrt{3}} \right). \]
This corresponds to an angle in the fourth quadrant. The angle whose tangent is \( -\frac{1}{\sqrt{3}} \) is \( -30^\circ \) (or \( 330^\circ \) in standard position).

### Step 3: Write in Polar Form
Now we can express \( \sqrt{3} - i \) in polar form:
\[ \sqrt{3} - i = 2 \left( \cos(-30^\circ) + i \sin(-30^\circ) \right) = 2 e^{-i \frac{\pi}{6}}. \]

### Step 4: Raise to the Power of 25
Now we raise \( \sqrt{3} - i \) to the power of 25:
\[ (\sqrt{3} - i)^{25} = (2 e^{-i \frac{\pi}{6}})^{25} = 2^{25} e^{-i \frac{25\pi}{6}}. \]

### Step 5: Find the Reciprocal
Now, we need to find the reciprocal:
\[ \frac{1}{(\sqrt{3} - i)^{25}} = \frac{1}{2^{25}} e^{i \frac{25\pi}{6}}. \]

### Step 6: Simplify the Argument
Next, we simplify the argument \( \frac{25\pi}{6} \):
\[ \frac{25\pi}{6} = 4\pi + \frac{\pi}{6} = 2(2\pi) + \frac{\pi}{6} \quad \text{(since } 4\pi \text{ is a full rotation)}. \]
Thus, the angle can be simplified to \( \frac{\pi}{6} \).

### Step 7: Conclusion
The angle that the vector representing the complex number \( \frac{1}{(\sqrt{3} - i)^{25}} \) makes with the positive direction of the real axis is:
\[ \frac{\pi}{6} \text{ or } 30^\circ. \]

### Final Answer
Thus, the angle is \( 30^\circ \).

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