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To solve the given equation:

\[
2z^4 + (1 - \sqrt{3}i) = 0
\]

we need to find the roots \( z \). Let's proceed step by step:

### Step 1: Simplify the Equation

Rearrange the equation to isolate \( z^4 \):

\[
2z^4 = - (1 - \sqrt{3}i)
\]

Divide by 2 on both sides:

\[
z^4 = \frac{-(1 - \sqrt{3}i)}{2}
\]

\[
z^4 = \frac{-1 + \sqrt{3}i}{2}
\]

### Step 2: Express the Right Side in Polar Form

The complex number \(\frac{-1 + \sqrt{3}i}{2}\) can be expressed in polar form as:

1. **Magnitude**:
   \[
   |r| = \sqrt{\left(\frac{-1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1
   \]

2. **Argument**:
   Since \(-1/2\) is the real part and \(\sqrt{3}/2\) is the imaginary part, the argument (angle) \(\theta\) is:
   \[
   \theta = \arctan\left(\frac{\sqrt{3}/2}{-1/2}\right) = \arctan(-\sqrt{3}) = \frac{2\pi}{3} \, (\text{as the point is in the second quadrant})
   \]

Thus, we can write:
\[
\frac{-1 + \sqrt{3}i}{2} = \text{cis}\left(\frac{2\pi}{3}\right) \quad \text{(where \(\text{cis}(\theta) = \cos(\theta) + i\sin(\theta)\))}
\]

### Step 3: Find the Fourth Roots Using De Moivre's Theorem

De Moivre's Theorem tells us that if \( z^4 = r\text{cis}(\theta) \), then:

\[
z = \sqrt[4]{r}\text{cis}\left(\frac{\theta + 2k\pi}{4}\right), \, k = 0, 1, 2, 3
\]

Since \( r = 1 \), we have \( \sqrt[4]{r} = 1 \). Now, we need to find the angles:

\[
z = \text{cis}\left(\frac{\frac{2\pi}{3} + 2k\pi}{4}\right)
\]

\[
z = \text{cis}\left(\frac{2\pi}{3} + 2k\pi \right) / 4
\]

### Step 4: Calculate the Four Roots

We evaluate \( k = 0, 1, 2, 3 \) to find the four solutions:

1. **For \( k = 0 \)**:
   \[
   z_1 = \text{cis}\left(\frac{2\pi}{12}\right) = \text{cis}\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2}i
   \]

2. **For \( k = 1 \)**:
   \[
   z_2 = \text{cis}\left(\frac{8\pi}{12}\right) = \text{cis}\left(\frac{2\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + \frac{\sqrt{3}}{2}i
   \]

3. **For \( k = 2 \)**:
   \[
   z_3 = \text{cis}\left(\frac{14\pi}{12}\right) = \text{cis}\left(\frac{7\pi}{6}\right) = \cos\left(\frac{7\pi}{6}\right) + i\sin\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} - \frac{1}{2}i
   \]

4. **For \( k = 3 \)**:
   \[
   z_4 = \text{cis}\left(\frac{20\pi}{12}\right) = \text{cis}\left(\frac{5\pi}{3}\right) = \cos\left(\frac{5\pi}{3}\right) + i\sin\left(\frac{5\pi}{3}\right) = \frac{1}{2} - \frac{\sqrt{3}}{2}i
   \]

### Final Solutions:

\[
z_1 = \frac{\sqrt{3}}{2} + \frac{1}{2}i, \; z_2 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \; z_3 = -\frac{\sqrt{3}}{2} - \frac{1}{2}i, \; z_4 = \frac{1}{2} - \frac{\sqrt{3}}{2}i
\]

Would you like a detailed breakdown of any step or further explanation?

### Related Questions:

1. What is De Moivre's Theorem and how is it applied to complex roots?
2. How do you convert a complex number from Cartesian form to polar form?
3. What are the characteristics of complex roots for polynomial equations?
4. How do magnitudes and angles affect the solutions of \( z^n = r \text{cis}(\theta) \)?
5. What is the geometric interpretation of these roots on the complex plane?

### Tip:
When working with complex numbers, converting to polar form often simplifies multiplication, division, and finding roots, making the calculations more manageable.

Find all roots (solutions) of the equation 2z^4 + (1 - √3i) = 0

Find the roots of the equation 2z^4 + (1 - √3i) = 0 using De Moivre's Theorem. This problem involves converting complex numbers to polar form, applying the theorem to find fourth roots, and calculating all solutions. Learn step-by-step how to solve this complex analysis problem.

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