To solve the equation $z^2 - 3 + 4i = 0$, we can follow these steps:

1. Rewrite the equation: $z^2 = 3 - 4i$

2. Express $z$ in polar form: Let $z = a + bi$. Therefore, $z^2 = (a + bi)^2 = a^2 - b^2 + 2abi$.

   Comparing the real and imaginary parts: $a^2 - b^2 = 3 \quad \text{(1)}$ $2ab = -4 \quad \text{(2)}$

3. Solve the system of equations: From equation (2), solve for $b$: $b = \frac{-4}{2a} = \frac{-2}{a}$

   Substitute $b$ into equation (1): $a^2 - \left(\frac{-2}{a}\right)^2 = 3$ $a^2 - \frac{4}{a^2} = 3$ Multiply through by $a^2$ to clear the fraction: $a^4 - 4 = 3a^2$ Rearrange into a standard form quadratic in $a^2$: $a^4 - 3a^2 - 4 = 0$ Let $u = a^2$: $u^2 - 3u - 4 = 0$

4. Solve the quadratic equation: Use the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 1$, $b = -3$, and $c = -4$: $u = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}$ Therefore, $u = 4$ or $u = -1$.

   Since $u = a^2$ and $a^2$ must be non-negative: $a^2 = 4 \implies a = \pm 2$

5. Find $b$ using $b = \frac{-2}{a}$: $\text{If } a = 2, \quad b = \frac{-2}{2} = -1$ $\text{If } a = -2, \quad b = \frac{-2}{-2} = 1$

6. Write the solutions for $z$: Therefore, the solutions are: $z = 2 - i \quad \text{and} \quad z = -2 + i$

So the solutions to the equation $z^2 - 3 + 4i = 0$ are $\boxed{2 - i \text{ and } -2 + i}$.

Do you want details or have any questions?

Here are five questions you might ask:

1. Can you explain the quadratic formula in more detail?
2. How do you convert complex numbers into polar form?
3. What are the properties of complex numbers?
4. How do you solve quadratic equations with complex coefficients?
5. What is the geometric interpretation of complex roots?

Tip: Always double-check your solutions by plugging them back into the original equation to ensure they satisfy it.