To solve this problem, we need to determine both the modulus $r$ and the argument $\theta$ of the complex expression $z^3 w^5$.

Step 1: Simplifying the expression for $z^3 w^5$

Given:

• $z \cdot \bar{z} = -1 + i$
• $w \cdot \bar{w} = -1 + \sqrt{3} i$

We need to first express $z$ and $w$ in polar form to make it easier to compute powers and products.

Step 2: Finding $z$

The relation $z \cdot \bar{z} = -1 + i$ tells us that $z \cdot \bar{z}$ is the modulus squared of $z$, which is equal to: $|z|^2 = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$ Therefore, the modulus of $z$ is $|z| = \sqrt[4]{2}$.

Next, to determine the argument of $z$, we take the argument of the complex number $-1 + i$. The argument $\theta_z$ of $-1 + i$ is: $\theta_z = \text{arg}(-1 + i) = \tan^{-1} \left( \frac{1}{-1} \right) = \frac{3\pi}{4}$

Thus, we can express $z$ as: $z = \sqrt[4]{2} e^{i \frac{3\pi}{4}}$

Step 3: Finding $w$

For $w$, the relation $w \cdot \bar{w} = -1 + \sqrt{3} i$ implies that the modulus squared of $w$ is: $|w|^2 = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$ So the modulus of $w$ is $|w| = \sqrt[4]{2}$.

To find the argument of $w$, we use the argument of the complex number $-1 + \sqrt{3} i$. This gives: $\theta_w = \text{arg}(-1 + \sqrt{3} i) = \tan^{-1} \left( \frac{\sqrt{3}}{-1} \right) = \frac{2\pi}{3}$

Thus, we can express $w$ as: $w = \sqrt[4]{2} e^{i \frac{2\pi}{3}}$

Step 4: Calculating $z^3 w^5$

Now we compute the modulus and argument of $z^3 w^5$. Using the property of exponentiation in polar form:

1. Modulus: [ |z^3 w^5| = |z|^3 \cdot |w|^5 = (\sqrt[4]{2})^3 \cdot (\sqrt[4]{2})^5 = 2^{\frac{3}{4}} \cdot 2^{\frac{5}{4}} = 2^{2} = 4 ]

2. Argument: $\text{arg}(z^3 w^5) = 3 \cdot \frac{3\pi}{4} + 5 \cdot \frac{2\pi}{3} = \frac{9\pi}{4} + \frac{10\pi}{3} = \frac{27\pi}{12} + \frac{40\pi}{12} = \frac{67\pi}{12}$ We reduce the argument by subtracting $2\pi$ until it lies within $(-\pi, \pi]$. Subtracting $2\pi$ (or $\frac{24\pi}{12}$) multiple times: $\frac{67\pi}{12} - 2\pi \times 5 = \frac{67\pi}{12} - \frac{60\pi}{12} = \frac{7\pi}{12}$ Thus, the argument is $\theta = \frac{7\pi}{12}$.

Final Answer:

• $r = 4$
• $\theta = \frac{7\pi}{12}$

Would you like more details or have any questions?

Here are 5 relative questions:

1. How can we convert a complex number into polar form?
2. What is the general method to calculate the modulus of a product of complex numbers?
3. How do you find the argument of a complex number in the complex plane?
4. What steps are involved in reducing an argument to be within $(-\pi, \pi]$?
5. Why is polar form useful in calculating powers of complex numbers?

Tip: When multiplying complex numbers, converting them into polar form makes it easier to handle both modulus and argument calculations.