https://file.aiquickdraw.com/mathbot/file/1727031150123p9woc28e.jpg

Let's break down the given problems one by one.

**(a) \( (i)^{1/3} \)**

To solve \( (i)^{1/3} \), we need to express \( i \) in polar form:

\[
i = e^{i\pi/2}
\]

Now, taking the cube root:

\[
(i)^{1/3} = \left( e^{i\pi/2} \right)^{1/3} = e^{i\pi/6}
\]

But since we are dealing with complex numbers, we must account for all possible roots by adding \( 2k\pi \) (where \( k \in \mathbb{Z} \)) to the argument:

\[
(i)^{1/3} = e^{i(\pi/6 + 2k\pi/3)}, \quad k = 0, 1, 2
\]

Thus, the three roots are:

\[
z_1 = e^{i\pi/6}, \quad z_2 = e^{i(5\pi/6)}, \quad z_3 = e^{i(9\pi/6)} = e^{i3\pi/2}
\]

**(b) \( (-4 + 3i)^{1/5} \)**

To find the fifth roots of \( -4 + 3i \), we first express it in polar form.

\[
-4 + 3i = r \text{cis} \theta
\]

where \( r = \sqrt{(-4)^2 + 3^2} = 5 \) and \( \theta = \tan^{-1} \left( \frac{3}{-4} \right) \approx 143.13^\circ \text{ or } \frac{3\pi}{4} \text{ radians}.
\)

The fifth roots are:

\[
(-4 + 3i)^{1/5} = 5^{1/5} \text{cis} \left( \frac{3\pi}{4 \cdot 5} + \frac{2k\pi}{5} \right), \quad k = 0, 1, 2, 3, 4
\]

The magnitude \( r^{1/5} = 5^{1/5} \approx 1.3797 \).

The arguments for the five roots will be:

\[
\theta_k = \frac{3\pi}{20} + \frac{2k\pi}{5}, \quad k = 0, 1, 2, 3, 4
\]

Thus, the roots are:

\[
z_k = 1.3797 \text{cis} \left( \frac{3\pi}{20} + \frac{2k\pi}{5} \right), \quad k = 0, 1, 2, 3, 4
\]

**(c) Solve \( z^2 + (i - 2)z + 3 - i = 0 \)**

This is a quadratic equation in \( z \). We can solve it using the quadratic formula:

\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the given quadratic:

\[
a = 1, \quad b = i - 2, \quad c = 3 - i
\]

Substituting into the quadratic formula:

\[
z = \frac{-(i-2) \pm \sqrt{(i-2)^2 - 4(1)(3-i)}}{2(1)}
\]

Now, simplifying the discriminant:

\[
(i-2)^2 = (i^2 - 4i + 4) = -1 - 4i + 4 = 3 - 4i
\]
\[
4(3 - i) = 12 - 4i
\]
\[
\text{Discriminant} = 3 - 4i - (12 - 4i) = 3 - 4i - 12 + 4i = -9
\]

So, the roots are:

\[
z = \frac{-(-2 + i) \pm \sqrt{-9}}{2} = \frac{2 - i \pm 3i}{2}
\]

The two roots are:

\[
z_1 = \frac{2 - i + 3i}{2} = \frac{2 + 2i}{2} = 1 + i
\]
\[
z_2 = \frac{2 - i - 3i}{2} = \frac{2 - 4i}{2} = 1 - 2i
\]

### Plotting

You can plot the roots of (a), (b), and (c) on the complex plane. Each solution gives a set of complex numbers that can be represented as points where the real part is on the x-axis and the imaginary part is on the y-axis.

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Would you like a detailed explanation or help with visualizing the plots?

Here are five related questions to deepen your understanding:
1. How do you determine the principal argument of a complex number?
2. What is the geometric interpretation of complex roots on the complex plane?
3. Why are there multiple solutions for roots of complex numbers?
4. How does the quadratic formula adapt when solving for complex numbers?
5. How do the polar coordinates of a complex number affect its roots?

**Tip**: When finding complex roots, always remember to consider all possible arguments by adding \( 2k\pi \) for each integer \( k \) within the root's range.

Find all values of the following roots and plot them: (a) (i)^(1/3), (b) (-4 + 3i)^(1/5), (c) z^2 + (i - 2)z + 3 - i = 0

Explore how to find the roots of complex numbers including (i)^(1/3), (-4 + 3i)^(1/5), and solving the quadratic equation z^2 + (i - 2)z + 3 - i = 0. This problem covers advanced topics such as De Moivre's Theorem and the Quadratic Formula, suitable for undergraduate students.

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