We want to solve the equation \( z^4 + 1 = j\sqrt{3} \) for all values of \( z \).

### Step 1: Rewrite the equation in standard form
The given equation can be rearranged as:

\[
z^4 = j\sqrt{3} - 1
\]

### Step 2: Convert the right-hand side to polar form
Let's convert the complex number \( j\sqrt{3} - 1 \) to its polar form. The complex number can be written as:

\[
j\sqrt{3} - 1 = -1 + j\sqrt{3}
\]

This is the same as the complex number we worked with earlier, so the polar form is:

\[
-1 + j\sqrt{3} = 2\left(\cos\frac{2\pi}{3} + j\sin\frac{2\pi}{3}\right)
\]

Thus:

\[
z^4 = 2\left(\cos\frac{2\pi}{3} + j\sin\frac{2\pi}{3}\right)
\]

### Step 3: Apply De Moivre's Theorem
We need to find the fourth roots of the complex number \( 2\left(\cos\frac{2\pi}{3} + j\sin\frac{2\pi}{3}\right) \). Let's express the equation as:

\[
z = \sqrt[4]{2} \left(\cos\frac{\frac{2\pi}{3} + 2k\pi}{4} + j\sin\frac{\frac{2\pi}{3} + 2k\pi}{4}\right)
\]

for \( k = 0, 1, 2, 3 \).

### Step 4: Calculate all four roots
The magnitude is:

\[
\sqrt[4]{2} = 2^{1/4}
\]

Now, calculate the angles for each \( k \):

1. **For \( k = 0 \):**
   \[
   z_0 = 2^{1/4}\left(\cos\frac{2\pi}{12} + j\sin\frac{2\pi}{12}\right) = 2^{1/4}\left(\cos\frac{\pi}{6} + j\sin\frac{\pi}{6}\right)
   \]
   \[
   z_0 = 2^{1/4}\left(\frac{\sqrt{3}}{2} + j\frac{1}{2}\right)
   \]

2. **For \( k = 1 \):**
   \[
   z_1 = 2^{1/4}\left(\cos\frac{2\pi + 8\pi}{12} + j\sin\frac{2\pi + 8\pi}{12}\right) = 2^{1/4}\left(\cos\frac{5\pi}{6} + j\sin\frac{5\pi}{6}\right)
   \]
   \[
   z_1 = 2^{1/4}\left(-\frac{\sqrt{3}}{2} + j\frac{1}{2}\right)
   \]

3. **For \( k = 2 \):**
   \[
   z_2 = 2^{1/4}\left(\cos\frac{2\pi + 16\pi}{12} + j\sin\frac{2\pi + 16\pi}{12}\right) = 2^{1/4}\left(\cos\frac{9\pi}{6} + j\sin\frac{9\pi}{6}\right)
   \]
   \[
   z_2 = 2^{1/4}\left(-\frac{\sqrt{3}}{2} - j\frac{1}{2}\right)
   \]

4. **For \( k = 3 \):**
   \[
   z_3 = 2^{1/4}\left(\cos\frac{2\pi + 24\pi}{12} + j\sin\frac{2\pi + 24\pi}{12}\right) = 2^{1/4}\left(\cos\frac{13\pi}{6} + j\sin\frac{13\pi}{6}\right)
   \]
   \[
   z_3 = 2^{1/4}\left(\frac{\sqrt{3}}{2} - j\frac{1}{2}\right)
   \]

### Final Solutions:
The four solutions for \( z \) are:

\[
z_0 = 2^{1/4}\left(\frac{\sqrt{3}}{2} + j\frac{1}{2}\right)
\]
\[
z_1 = 2^{1/4}\left(-\frac{\sqrt{3}}{2} + j\frac{1}{2}\right)
\]
\[
z_2 = 2^{1/4}\left(-\frac{\sqrt{3}}{2} - j\frac{1}{2}\right)
\]
\[
z_3 = 2^{1/4}\left(\frac{\sqrt{3}}{2} - j\frac{1}{2}\right)
\]

These are the four complex roots of the equation \( z^4 + 1 = j\sqrt{3} \).

Would you like any further explanation or details on this solution?

### Related Questions:
1. How do you find the roots of a general polynomial equation in the complex plane?
2. What is the geometric significance of the roots of a complex number on the Argand diagram?
3. Can you explain De Moivre's Theorem in more detail?
4. How do you compute higher-order roots of a complex number?
5. What is the importance of the modulus and argument in solving complex equations?

### Tip:
When solving for complex roots, remember that each root corresponds to a rotation on the complex plane, evenly distributed around the origin.

Learn how to solve the equation z^4 + 1 = j√3 for all values of z using complex numbers and De Moivre's Theorem. Explore the polar form of complex numbers and find the fourth roots of a complex number. Suitable for advanced high school and college students.

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