The equation shown in the image is $z^2 = 1 + i$, where $z$ is a complex number, and $i$ is the imaginary unit. This is a quadratic equation in the complex number $z$.

To determine how many solutions this equation has, we need to solve it. We start by expressing the right-hand side of the equation in polar form.

1. Convert $1 + i$ into polar form: $1 + i = \sqrt{1^2 + 1^2} \left( \cos \theta + i \sin \theta \right) = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$ So, the complex number $1 + i$ can be written as $\sqrt{2} e^{i \frac{\pi}{4}}$.

2. Solve $z^2 = \sqrt{2} e^{i \frac{\pi}{4}}$.

For a complex number $z$ such that $z^2 = re^{i\theta}$, we know the general solutions are given by: $z = \sqrt{r} e^{i\frac{\theta}{2}} \quad \text{or} \quad z = \sqrt{r} e^{i\left(\frac{\theta}{2} + \pi\right)}$

Since $z^2 = \sqrt{2} e^{i \frac{\pi}{4}}$, we apply this formula:

• $r = \sqrt{2}$, so $\sqrt{r} = \sqrt{\sqrt{2}} = 2^{1/4}$.
• $\theta = \frac{\pi}{4}$, so the two solutions for $z$ will be:
  1. $z_1 = 2^{1/4} e^{i \frac{\pi}{8}}$
  2. $z_2 = 2^{1/4} e^{i \left( \frac{\pi}{8} + \pi \right)} = 2^{1/4} e^{i \frac{9\pi}{8}}$

Therefore, the quadratic equation $z^2 = 1 + i$ has 2 solutions.

Would you like a more detailed explanation of the process, or do you have any questions?

Here are some related questions for you:

1. How do you convert complex numbers to polar form?
2. What is the significance of the complex square root in polar form?
3. Can this method be extended to higher powers of complex numbers?
4. How do you solve for complex numbers in higher-order equations (e.g., $z^n = 1 + i$)?
5. What is the general form of the solution for $z^n = r e^{i \theta}$?

Tip: When solving complex equations, always check for the number of solutions based on the degree of the equation!