What is the square root of i, where `i=sqrt(-1)` ?

Options

A. `(1+i)/(2)`
B. `(1-i)/(2)`
C. `(1+i)/(sqrt2)`
D. None of these

Explanation

To find the square root of \( i \) (where \( i = \sqrt{-1} \)), we can express \( i \) in polar form and then apply the properties of complex numbers. Here’s a step-by-step solution:

### Step 1: Express \( i \) in polar form
The complex number \( i \) can be represented in polar form as:
\[ i = 0 + 1i \]
In polar form, this can be expressed as:
\[ i = r(\cos \theta + i \sin \theta) \]
where \( r = |i| \) (the modulus) and \( \theta \) (the argument).

### Step 2: Calculate the modulus \( r \)
The modulus \( r \) of \( i \) is given by:
\[ r = |i| = \sqrt{0^2 + 1^2} = \sqrt{1} = 1 \]

### Step 3: Determine the argument \( \theta \)
The argument \( \theta \) of \( i \) is:
\[ \theta = \frac{\pi}{2} \quad (\text{since } i \text{ lies on the positive imaginary axis}) \]

### Step 4: Write \( i \) in polar form
Now we can write \( i \) in polar form:
\[ i = 1 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) \]

### Step 5: Find the square root of \( i \)
To find the square root of \( i \), we use the formula for the square root of a complex number in polar form:
\[ \sqrt{r} \left( \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right) \]
Substituting \( r = 1 \) and \( \theta = \frac{\pi}{2} \):
\[ \sqrt{i} = \sqrt{1} \left( \cos \frac{\pi/2}{2} + i \sin \frac{\pi/2}{2} \right) = 1 \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \]

### Step 6: Calculate \( \cos \frac{\pi}{4} \) and \( \sin \frac{\pi}{4} \)
We know:
\[ \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \]
Thus,
\[ \sqrt{i} = 1 \left( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \]

### Step 7: Final result
Therefore, the square root of \( i \) is:
\[ \sqrt{i} = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \]