The equation given is:

\[
-z^2 = 2 + i
\]

To solve for \( z \), follow these steps:

### Step 1: Rewrite the equation
First, express the equation as:

\[
z^2 = -(2 + i)
\]
This simplifies to:
\[
z^2 = -2 - i
\]

Now, we need to find \( z \) such that \( z^2 = -2 - i \).

### Step 2: Express \( z \) in polar form
We will express \( z = x + iy \) (where \( x \) and \( y \) are real numbers) in polar form to simplify the solution.

First, express the right-hand side \( -2 - i \) in polar form.

1. Find the modulus \( r \) of \( -2 - i \):
\[
r = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}
\]

2. Find the argument \( \theta \) (the angle in the complex plane):
\[
\theta = \text{atan2}(-1, -2) = \pi + \text{atan2}(1, 2) \approx \pi + 0.4636 = 3.6052 \text{ radians}
\]

Thus, \( -2 - i \) in polar form is:
\[
-2 - i = \sqrt{5} \left( \cos(3.6052) + i \sin(3.6052) \right)
\]

### Step 3: Solve for \( z \)
To solve \( z^2 = \sqrt{5} \left( \cos(3.6052) + i \sin(3.6052) \right) \), take the square root of both sides.

The square roots of a complex number \( re^{i\theta} \) are given by:
\[
z = \sqrt{r} \left( \cos\left(\frac{\theta}{2}\right) + i \sin

Learn how to solve the complex quadratic equation -z^2 = 2 + i by converting complex numbers into polar form. Follow a detailed step-by-step guide covering the square root of complex numbers, polar representation, and De Moivre's theorem. Suitable for advanced high school or undergraduate students.

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