`"The value of"`: `i^(1//3)` is

Options

A. `(sqrt3+i)/(2)`
B. `(sqrt3-i)/2`
C. `(1+isqrt3)/2`
D. `(1-isqrt3)/2`

Answer

To find the value of \( i^{1/3} \), we can follow these steps:

### Step 1: Express \( i \) in exponential form
The complex number \( i \) can be expressed in polar form as:
\[ i = e^{i\frac{\pi}{2}} \]
This is because \( i \) is located at an angle of \( \frac{\pi}{2} \) radians on the complex plane.

### Step 2: Apply the exponent
Now, we need to find \( i^{1/3} \):
\[ i^{1/3} = \left(e^{i\frac{\pi}{2}}\right)^{1/3} \]
Using the property of exponents, we can simplify this to:
\[ i^{1/3} = e^{i\frac{\pi}{6}} \]

### Step 3: Find the principal value
The principal value of \( e^{i\frac{\pi}{6}} \) can be expressed in rectangular form:
\[ e^{i\frac{\pi}{6}} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) \]
Calculating the cosine and sine values:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Thus,
\[ e^{i\frac{\pi}{6}} = \frac{\sqrt{3}}{2} + i\frac{1}{2} \]

### Step 4: Consider other cube roots
Since we are looking for the cube roots, we need to consider all possible cube roots. The general formula for the \( n \)-th roots of a complex number is given by:
\[ z_k = r^{1/n} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i \sin\left(\frac{\theta + 2k\pi}{n}\right) \right) \]
where \( k = 0, 1, 2, \ldots, n-1 \).

For our case, \( n = 3 \):
1. For \( k = 0 \):
\[ z_0 = e^{i\frac{\pi}{6}} = \frac{\sqrt{3}}{2} + i\frac{1}{2} \]
2. For \( k = 1 \):
\[ z_1 = e^{i\left(\frac{\pi}{6} + \frac{2\pi}{3}\right)} = e^{i\frac{5\pi}{6}} = -\frac{\sqrt{3}}{2} + i\frac{1}{2} \]
3. For \( k = 2 \):
\[ z_2 = e^{i\left(\frac{\pi}{6} + \frac{4\pi}{3}\right)} = e^{i\frac{3\pi}{2}} = -i \]

### Final Result
The three cube roots of \( i \) are:
\[ \frac{\sqrt{3}}{2} + i\frac{1}{2}, \quad -\frac{\sqrt{3}}{2} + i\frac{1}{2}, \quad -i \]