Let's solve the problem given in the image:

You are provided with the complex numbers $z = -1 + 3i$ and $w = 3 - i$.

Part (a) $\left| \bar{z} w \right|$

Here, $\bar{z}$ is the complex conjugate of $z$. The complex conjugate of $z = -1 + 3i$ is $\bar{z} = -1 - 3i$.

Now, we want to compute $\left| \bar{z} w \right|$.

1. First, find the product $\bar{z} w$: $\bar{z} \cdot w = (-1 - 3i)(3 - i)$ Using the distributive property: = -3 + i - 9i + 3i^2$$ Since $$i^2 = -1$$, we substitute $$3i^2 = 3(-1) = -3$$: $$= -3 + i - 9i - 3 = -6 - 8i$$
2. Now, compute the modulus $\left| \bar{z} w \right|$. For a complex number $a + bi$, the modulus is: $\left| a + bi \right| = \sqrt{a^2 + b^2}$ So, for $-6 - 8i$: $\left| -6 - 8i \right| = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Thus, $\left| \bar{z} w \right| = 10$.

Part (b) $\text{Im}\left( \frac{w}{z} \right)$

We need to find the imaginary part of $\frac{w}{z}$.

1. First, compute $\frac{w}{z}$. This is done by multiplying the numerator and the denominator by the complex conjugate of the denominator: $\frac{w}{z} = \frac{3 - i}{-1 + 3i} \times \frac{-1 - 3i}{-1 - 3i}$ Simplify the numerator: = -3 - 9i + i + 3i^2 = -3 - 9i + i - 3 = -6 - 8i$$ Simplify the denominator: $$(-1 + 3i)(-1 - 3i) = (-1)^2 - (3i)^2 = 1 - 9(-1) = 1 + 9 = 10$$ So, $$\frac{w}{z} = \frac{-6 - 8i}{10} = -\frac{6}{10} - \frac{8}{10}i = -0.6 - 0.8i$$.
2. The imaginary part of $-0.6 - 0.8i$ is $-0.8$.

Thus, $\text{Im}\left( \frac{w}{z} \right) = -0.8$.

Let me know if you need more details or have any questions!

Here are five related questions:

1. How do you find the modulus of a complex number?
2. What is the process for dividing complex numbers?
3. What is the geometric meaning of the complex conjugate?
4. Can you explain how to simplify the expression when multiplying complex numbers?
5. How does the imaginary part affect the properties of a complex number?

Tip: Always express complex division in terms of multiplying by the conjugate to simplify the calculation.