To find all complex numbers $z$ such that $z^2 = 8 - 6i$, we will solve this by expressing the right-hand side in polar form and applying the square root to obtain $z$.

Step 1: Write $8 - 6i$ in polar form.

1. Magnitude: The magnitude $r$ of $8 - 6i$ is given by:

   $r = |8 - 6i| = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

2. Argument: The argument $\theta$ is the angle the complex number makes with the positive real axis, which is:

   $\theta = \tan^{-1} \left( \frac{-6}{8} \right) = \tan^{-1} \left( -\frac{3}{4} \right)$

   Using known values, we get:

   $\theta \approx -0.6435 \text{ radians} \quad \text{(approximately)}$

Thus, the polar form of $8 - 6i$ is:

$8 - 6i = 10 \text{cis}(-0.6435) \quad \text{where} \ \text{cis}(\theta) = \cos(\theta) + i\sin(\theta)$

Step 2: Solve for $z$ such that $z^2 = 8 - 6i$.

To solve $z^2 = 10 \text{cis}(-0.6435)$, we take the square root of both sides. This gives two solutions:

$z = \sqrt{10} \text{cis} \left( \frac{-0.6435 + 2k\pi}{2} \right) \quad \text{for} \ k = 0, 1$

The two solutions correspond to the two possible values of $k$.

Step 3: Find the two solutions.

For $k = 0$:

$z_1 = \sqrt{10} \text{cis} \left( \frac{-0.6435}{2} \right) = \sqrt{10} \text{cis}(-0.3218)$

For $k = 1$:

$z_2 = \sqrt{10} \text{cis} \left( \frac{-0.6435 + 2\pi}{2} \right) = \sqrt{10} \text{cis}(2.8198)$

Step 4: Convert back to rectangular form.

Now, convert the polar form back into rectangular form:

1. For $z_1$:

$z_1 = \sqrt{10} (\cos(-0.3218) + i\sin(-0.3218)) = \sqrt{10}(0.9499 - 0.3127i)$

Thus:

$z_1 \approx 3.0 - 1.0i$

2. For $z_2$:

$z_2 = \sqrt{10} (\cos(2.8198) + i\sin(2.8198)) = \sqrt{10}(-0.9499 + 0.3127i)$

Thus:

$z_2 \approx -3.0 + 1.0i$

Final Answer:

The two solutions are approximately:

$z_1 \approx 3.0 - 1.0i \quad \text{and} \quad z_2 \approx -3.0 + 1.0i$

Would you like more details or have any questions?

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Related questions:

1. How do we convert a complex number to polar form?
2. What is De Moivre's Theorem and how is it used?
3. How can we determine the argument of a complex number?
4. What is the geometric interpretation of taking the square root of a complex number?
5. How can we verify the solution by squaring $z$?

Tip: When finding square roots of complex numbers, it's often helpful to convert them into polar form for easier manipulation.