If `sqrt(3)+i= r(cos theta+isin theta)`, then find the value of `theta` in radian measure.

Answer

To solve the problem, we need to express the complex number \( \sqrt{3} + i \) in polar form and find the angle \( \theta \).

### Step-by-Step Solution:

1. **Identify the complex number**:
The given complex number is \( z = \sqrt{3} + i \).

2. **Calculate the modulus \( r \)**:
The modulus \( r \) of a complex number \( z = a + bi \) is given by:
\[ r = \sqrt{a^2 + b^2} \]
Here, \( a = \sqrt{3} \) and \( b = 1 \).
\[ r = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

3. **Express in polar form**:
We can express \( z \) in polar form as:
\[ z = r(\cos \theta + i \sin \theta) \]
Substituting the value of \( r \):
\[ z = 2(\cos \theta + i \sin \theta) \]

4. **Find \( \cos \theta \) and \( \sin \theta \)**:
From the polar form, we can equate:
\[ \cos \theta = \frac{\text{Re}(z)}{r} = \frac{\sqrt{3}}{2} \]
\[ \sin \theta = \frac{\text{Im}(z)}{r} = \frac{1}{2} \]

5. **Determine \( \theta \)**:
We know that:
- \( \cos \theta = \frac{\sqrt{3}}{2} \) corresponds to \( \theta = \frac{\pi}{6} \) (or 30 degrees).
- \( \sin \theta = \frac{1}{2} \) also corresponds to \( \theta = \frac{\pi}{6} \) (or 30 degrees).

Since both sine and cosine are positive, \( \theta \) is in the first quadrant.

6. **Final answer**:
Therefore, the value of \( \theta \) in radian measure is:
\[ \theta = \frac{\pi}{6} \]