The modulus -amplitude from of `sqrt3+i, where` ` i =sqrt(-1)` is

Options

A. `2(cos""(pi)/(3)+isin""(pi)/(3))`
B. `2(cos""(pi)/(6)+isin""(pi)/(6))`
C. `2(cos""(pi)/(3)-isin""(pi)/(3))`
D. `2(cos""(pi)/(6)-isin""(pi)/(6))`

Explanation

To find the modulus-amplitude form of the complex number \( z = \sqrt{3} + i \), we will follow these steps:

### Step 1: Identify the components of the complex number
The complex number can be expressed in the form \( z = a + bi \), where \( a = \sqrt{3} \) and \( b = 1 \).

### Step 2: Calculate the modulus \( r \)
The modulus \( r \) of a complex number \( z = a + bi \) is given by the formula:
\[ r = \sqrt{a^2 + b^2} \]
Substituting the values of \( a \) and \( b \):
\[ r = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

### Step 3: Calculate the argument \( \theta \)
The argument \( \theta \) can be found using the formula:
\[ \tan \theta = \frac{b}{a} \]
Substituting the values of \( a \) and \( b \):
\[ \tan \theta = \frac{1}{\sqrt{3}} \]
To find \( \theta \), we take the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]
We know that \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \), so:
\[ \theta = \frac{\pi}{6} \]

### Step 4: Write the modulus-amplitude form
The modulus-amplitude form of a complex number is given by:
\[ z = r(\cos \theta + i \sin \theta) \]
Substituting the values of \( r \) and \( \theta \):
\[ z = 2\left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right) \]

### Final Answer
Thus, the modulus-amplitude form of \( \sqrt{3} + i \) is:
\[ z = 2\left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right) \]

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