Apply De Moivre's Theorem to solve the equation $$ z^{4}=1 $$, thus obtaining the fourth roots of 1 .

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To find the fourth roots of $$ 1 $$ using **De Moivre's Theorem**, follow these steps:

### Step 1: Express $$ 1 $$ in Polar Form
The complex number $$ 1 $$ can be expressed in polar form as:
$$
1 = \cos(0) + i\sin(0) = \text{cis}(0)
$$
More generally, since angles are periodic with period $$ 2\pi $$, we can write:
$$
1 = \text{cis}(2k\pi) \quad \text{for any integer } k
$$

### Step 2: Apply De Moivre's Theorem
De Moivre's Theorem states that for any complex number $$ z = \text{cis}(\theta) $$, the $$ n $$-th roots of $$ z $$ are given by:
$$
z^{1/n} = \text{cis}\left(\frac{\theta + 2k\pi}{n}\right) \quad \text{for } k = 0, 1, 2, \dots, n-1
$$
Here, $$ n = 4 $$ since we are finding the fourth roots.

### Step 3: Calculate the Fourth Roots
Substituting $$ \theta = 0 $$ (from $$ 1 = \text{cis}(0) $$) and $$ n = 4 $$:
$$
z = \text{cis}\left(\frac{0 + 2k\pi}{4}\right) = \text{cis}\left(\frac{k\pi}{2}\right) \quad \text{for } k = 0, 1, 2, 3
$$
Calculating for each $$ k $$:
$$
\begin{align*}
k = 0: & \quad \text{cis}(0) = 1 \
k = 1: & \quad \text{cis}\left(\frac{\pi}{2}\right) = i \
k = 2: & \quad \text{cis}(\pi) = -1 \
k = 3: & \quad \text{cis}\left(\frac{3\pi}{2}\right) = -i \
\end{align*}
$$

### Final Answer
The four fourth roots of $$ 1 $$ are:
$$
1, \quad i, \quad -1, \quad -i
$$

**Answer:** All four fourth roots of 1 are 1, –1, i, and –i.
The fourth roots of 1 are 1, –1, i, and –i.

Apply De Moivre's Theorem to solve the equation \( z^{4}=1 \), thus obtaining the fourth roots of 1 .

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