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.