Detailed Answer

a) To determine the form of each complex number:

• \(\text{a}\) is in exponential form.
• \(\text{b}\) is in polar form.
• \(\text{c}\) is in Cartesian form.
• \(\text{d}\) is in Cartesian form.
• \(\text{e}\) is in Cartesian form.
• \(\text{f}\) is in polar form.

b) For each number, the real part is:

• \(\text{a} = 5 \cos \left(\frac{\pi}{3}\right)\)
• \(\text{b} = 5 \cos \left(\frac{\pi}{3}\right)\)
• \(\text{c} = 4\)
• \(\text{d} = 4 \cos \left(\frac{3 \pi}{2}\right)\)
• \(\text{e} = 0\)
• \(\text{f} = 12 \cos \left(\frac{\pi}{6}\right)\)

c) For each number, the imaginary part is:

• \(\text{a} = 5 \sin \left(\frac{\pi}{3}\right)\)
• \(\text{b} = 5 \sin \left(\frac{\pi}{3}\right)\)
• \(\text{c} = 3\)
• \(\text{d} = \sin \left(\frac{\pi}{3}\right)\)
• \(\text{e} = 2 e^{3}\)
• \(\text{f} = 12 \sin \left(\frac{\pi}{6}\right)\)

d) The argument of each number is:

• \(\text{a} = \frac{\pi}{3}\)
• \(\text{b} = \frac{\pi}{3}\)
• \(\text{c} = \tan^{-1} \left(\frac{3}{4}\right)\)
• \(\text{d} = \frac{\pi}{2}\)
• \(\text{e} = \frac{\pi}{2} \)
• \(\text{f} = \frac{\pi}{6}\)

Detailed Answer

a) To calculate \( z_{1} \times z_{2} \):

\[ z_{1} \times z_{2} = (3 + 4i) \times (1 - 3i) = 3 \times 1 - 3 \times 3i + 4i \times 1 - 4i \times 3i \]

\[ = 3 - 9i + 4i + 12 \]

\[ = 15 - 5i \]

b) To calculate \( \frac{z_{1}}{z_{2}} \):

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

\[ \frac{z_{1}}{z_{2}} \times \frac{1 + 3i}{1 + 3i} = \frac{(3 + 4i)(1 + 3i)}{(1 - 3i)(1 + 3i)} \]

\[ = \frac{3 + 9i + 4i + 12i^2}{1 - 9i^2} \]

\[ = \frac{3 + 13i - 12}{1 + 9} = \frac{-9 + 13i}{10} \]

c) To calculate \( z_{2}^{2} \):

\[ z_{2}^{2} = (1 - 3i)^{2} \]

\[ = 1 - 6i + 9i^{2} \]

\[ = 1 - 6i - 9 \]

\[ = -8 - 6i \]

d) To convert \( z_{1} \) and \( z_{2} \) to exponential form:

\[ z_{1} = 3 + 4i \]

\[ |z_{1}| = \sqrt{3^2 + 4^2} = 5 \]

\[ \theta_{1} = \tan^{-1}\left(\frac{4}{3}\right) \]

\[ z_{1} = 5e^{iarctan(\frac{4}{3})} \]

\[ z_{2} = 1 - 3i \]

\[ |z_{2}| = \sqrt{1^2 + (-3)^2} = \sqrt{10} \]

\[ \theta_{2} = \tan^{-1}\left(-3\right) \]

\[ z_{2} = \sqrt{10}e^{-iarctan(3)} \]

e) To calculate \( z_{1} \times z_{2} \) and \( \frac{z_{1}}{z_{2}} \) with the exponential form:

\[ z_{1} \times z_{2} = \left(5e^{i \tan^{-1}\left(\frac{4}{3}\right)}\right) \times \left(\sqrt{10}e^{i \tan^{-1}\left(-3\right)}\right) \]

\[ = 5 \sqrt{10} e^{i \left(\tan^{-1}\left(\frac{4}{3}\right) - \tan^{-1}\left(3\right)\right)} \]

\[ \frac{z_{1}}{z_{2}} = \frac{5e^{i \tan^{-1}\left(\frac{4}{3}\right)}}{\sqrt{10}e^{i \tan^{-1}\left(-3\right)}} \]

\[ = \frac{5}{\sqrt{10}} e^{i \left(\tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(3\right)\right)} \]

f) The value of \( z_{2}^{2} \) in exponential form:

\[ z_{2}^{2} = (\sqrt{10}e^{-iarctan(3)})^2 = 10e^{-2iarctan(3)}\]

Detailed Answer

a) We are given the argument, and \( A \) is a circle of radius 2 around the origin.

\[ z_1 = 2e^{i\frac{\pi}{3}} \]

b) Same as in (a), although we have to be mindful of the negative sign as the angle goes clockwise:

\[ z_2 = 3e^{-i\frac{\pi}{6}} \]

c) To find \(\gamma\) and \(\left|z_{3}\right|\), we use the product of the complex numbers:

\[ z_3 = z_1 \times z_2 = \left(2e^{i\frac{\pi}{3}}\right) \times \left(3e^{-i\frac{\pi}{6}}\right) \]

\[ z_3 = 6e^{i\left(\frac{\pi}{3} - \frac{\pi}{6}\right)} = 6e^{i\frac{\pi}{6}} \]

The modulus of \( z_3 \) is \(\left|z_3\right| = 6\), and the argument \(\gamma\) is \(\frac{\pi}{6}\).

d) To express \( z_3 \) in Cartesian form:

\[ z_3 = 6e^{i\frac{\pi}{6}} = 6 \left(\cos\frac{\pi}{6} + i \sin\frac{\pi}{6}\right) \]

\[ \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \sin\frac{\pi}{6} = \frac{1}{2} \]

\[ z_3 = 6 \left(\frac{\sqrt{3}}{2} + i \frac{1}{2}\right) = 3\sqrt{3} + 3i \]

e) To show that \( \mathfrak{R}\left(z_{1}\right) + 2 \times \mathfrak{R}\left(z_{2}\right) = \mathfrak{R}\left(z_{3}\right) \):

\[ \mathfrak{R}(z_1) = 2 \cos\frac{\pi}{3} = 2 \times \frac{1}{2} = 1 \]

\[ \mathfrak{R}(z_2) = 3 \cos\left(-\frac{\pi}{6}\right) = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]

\[ 2 \times \mathfrak{R}(z_2) = 2 \times \frac{3\sqrt{3}}{2} = 3\sqrt{3} \]

\[ \mathfrak{R}(z_1) + 2 \times \mathfrak{R}(z_2) = 1 + 3\sqrt{3} \]

\[ \mathfrak{R}(z_3) = 3\sqrt{3} \]

Thus, we have shown that \( \mathfrak{R}(z_1) + 2 \times \mathfrak{R}(z_2) = \mathfrak{R}(z_3) + 1 \).

Detailed Answer

a)

\[ \left|z_1\right| = \sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.61 \]

\[ \left|z_2\right| = \sqrt{4^2 + 1^2} = \sqrt{17} \approx 4.12 \]

b)

\[ \alpha = \arctan\left(\frac{3}{2}\right) \approx 0.983 \text{ radians} \]

\[ \beta = \arctan\left(\frac{1}{4}\right) \approx 0.245 \text{ radians} \]

c)

\[ \text{Area} = \frac{1}{2} \times OA \times OB \times \sin(\alpha - \beta) \]

\[ \text{Area} = \frac{1}{2} \times 3.61 \times 4.12 \times \sin(0.983 - 0.245) \approx 5.00 \]

d)

\[ \overline{z_1} = 2 - 3i \]

\[ z_{2} + \overline{z_{1}} = (4 + i) + (2 - 3i) = 6 - 2i \]

So, the coordinates of \( C \) are \( (6, -2) \).

e)

\[ \arg(z_{2} + \overline{z_{1}}) = \arg(6 - 2i) \]

\[ \arg(6 - 2i) = \arctan\left(\frac{-2}{6}\right) \approx -0.322 \text{ radians} \]

f)

\[ \left|z_{2} + \overline{z_{1}}\right| = \sqrt{6^2 + (-2)^2} = \sqrt{40} = 2\sqrt{10}\]

Therefore:

\[ z_{2} + \overline{z_{1}} = 2\sqrt{10} e^{-i0.322} \]

Detailed Answer

a)

\[ V_1 = \mathfrak{R}(20e^{12it}) \]

\[ V_2 = \mathfrak{R}(10e^{12it+15}) \]

\[20e^{12it} + 10e^{12it+15} = 20e^{12it}(1+\frac{1}{2}e^{i15})\]

\[ 1+\frac{1}{2}e^{i15} = 0.700 \times e^{i0.483} \]

So:

\[20e^{12it} + 10e^{12it+15} = 14e^{i(12t+0.48)} \]

By taking the real part of this number we get:

\[ V = 14cos(12t + 0.48) \]

b) We use a similar approach, first by rewriting:

\[ V_3 = 5sin(12t) = -5cos(12t + \frac{\pi}{2}) \]

\[20e^{12it} - 5e^{12it+\frac{\pi}{2}} = 20e^{12it}(1-\frac{1}{4}e^{i\frac{\pi}{2}})\]

\[ 1-\frac{1}{4}e^{i\frac{\pi}{2}} = 1.0 \times e^{-0.25i} \]

\[ V_1 + V_3 = \mathfrak{R}(20e^{12it}(1.0 \times e^{-0.25i})) = \mathfrak{R}(21e^{i(12t-0.25)})\]

At this step be sure to use the exact value of 1.0:

\[ V_1 + V_3 = 21cos(12t-0.25) \]

Detailed Answer

a) We solve the quadratic equation \( x^2 - 4x + 8 = 0 \) using the quadratic formula:

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

Here, \( a = 1 \), \( b = -4 \), and \( c = 8 \). Substituting these values:

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(8)}}{2(1)} = \frac{4 \pm \sqrt{16 - 32}}{2} = \frac{4 \pm \sqrt{-16}}{2} \]

\[ x = \frac{4 \pm 4i}{2} = 2 \pm 2i \]

So, the solutions are \( x = 2 + 2i \) and \( x = 2 - 2i \).

b) Here, \( a = 2 \), \( b = 5 \), and \( c = 4 \). Substituting these values:

\[ x = \frac{-5 \pm \sqrt{5^2 - 4(2)(4)}}{2(2)} = \frac{-5 \pm \sqrt{25 - 32}}{4} = \frac{-5 \pm \sqrt{-7}}{4} \]

\[ x = \frac{-5 \pm \sqrt{7}i}{4} = -\frac{5}{4} \pm \frac{\sqrt{7}}{4}i \]

So, the solutions are \( x = -\frac{5}{4} + \frac{\sqrt{7}}{4}i \) and \( x = -\frac{5}{4} - \frac{\sqrt{7}}{4}i \).

c) We solve it by factoring:

\[ x(9x^2 + 2x + 1) = 0 \]

So, one solution is \( x = 0 \). To find the other solutions, solve the quadratic equation \( 9x^2 + 2x + 1 = 0 \) using the quadratic formula:

\[ x = \frac{-2 \pm \sqrt{2^2 - 4(9)(1)}}{2(9)} = \frac{-2 \pm \sqrt{4 - 36}}{18} = \frac{-2 \pm \sqrt{-32}}{18} \]

\[ x = \frac{-2 \pm 4i\sqrt{2}}{18} = -\frac{1}{9} \pm \frac{2\sqrt{2}}{9}i \]

So, the solutions are \( x = 0 \), \( x = -\frac{1}{9} + \frac{2\sqrt{2}}{9}i \), and \( x = -\frac{1}{9} - \frac{2\sqrt{2}}{9}i \).

Detailed Answer

a)

\[ z_3 = -5i - (4 + 2i) = -4 - 7i \]

b)

\[ z_3 = (4 + 2i) - (-5i) = 4 + 7i \]

c)

\[ z_3 = (2 + 4i) + (1 - i) = 3 + 3i \]

d)

\[ z_3 = (2 - 4i) + (1 - i) = 3 - 5i \]

e)

\[ z_3 = \overline{(2 + 4i) + (1 - i)} = \overline{3 + 3i} = 3 - 3i \]

f)

\[ z_3 = (2 + 4i)(2 - 4i) = 4 - 16i^2 = 4 + 16 = 20 \]

Detailed Answer

a) The complex number \( z = 1 + 2i \) corresponds to the point (1, 2) on the complex plane.

To find \( iz \), multiply \( z \) by \( i \):

\[ iz = i(1 + 2i) = i + 2i^2 = i - 2 = -2 + i \]

This corresponds to the point (-2, 1) on the complex plane.

b) By visual inspection, the angle between the vectors \( z \) and \( iz \) appears to be 90 degrees, as multiplying a complex number by \( i \) results in a rotation of 90 degrees counterclockwise on the complex plane.

c) For \( z = 1 + 2i \):

\[ \arg(z) = \arctan\left(\frac{2}{1}\right) \approx 63.43^\circ \]

For \( iz = -2 + i \) (the vector is in the second quadrant):

\[ \arg(iz) = \arctan\left(\frac{1}{-2}\right) = -26.57+180 \approx 153.43^\circ \]

d) The angle between the two vectors is indeed 90 degrees, as expected. Multiplying a complex number by \( i \) rotates it by 90 degrees counterclockwise on the complex plane. The difference between \( \arg(iz) \) and \( \arg(z) \) confirms this:

\[ 153.43 - 63.43 = 90^\circ \]

Detailed Answer

a) \( \text{arctan}(\frac{\sqrt{3}}{3}) = \frac{\pi}{6}\)

b) \( \text{arctan}(\frac{-25}{25}) = arctan(-1) = -\frac{\pi}{4}\)

c) \( \text{arctan}(\frac{\sqrt{3}}{3}) = \frac{\pi}{6}\)

d) 0, as this is a real number.

e) \( \frac{\pi}{2} \), as the number is strictly imaginary.

Detailed Answer

a)

\[ r = \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} \]

\[ \theta = \arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} \]

So, the polar form is \( 2\sqrt{3} \left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) \).

b)

\[ r = \sqrt{(-25)^2 + 25^2} = \sqrt{625 + 625} = \sqrt{1250} = 25\sqrt{2} \]

The argument \( \theta \) is:

\[ \theta = \arctan\left(\frac{25}{-25}\right) = \arctan(-1) = -\frac{\pi}{4} \]

So, the polar form is \( 25\sqrt{2} \left(\cos\frac{\pi}{4} - i\sin\frac{\pi}{4}\right) \).

c)

\[ r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

The argument \( \theta \) is:

\[ \theta = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]

So, the polar form is \( 2 \left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) \).

Detailed Answer

a) To express \( \frac{1}{(1 - 2i)^4} \) in the form \( \frac{a}{b} \), we will first compute \((1 - 2i)^4\) and then find its reciprocal.

Step 1: Calculate \((1 - 2i)^4\)

\[ (1 - 2i)^2 = 1^2 - 2 \cdot 1 \cdot 2i + (2i)^2 = 1 - 4i + 4i^2 \]

Since \(i^2 = -1\):

\[ 1 - 4i + 4(-1) = 1 - 4i - 4 = -3 - 4i \]

Next, compute \((1 - 2i)^4\) by squaring \((1 - 2i)^2\):

\[ (-3 - 4i)^2 = (-3)^2 - 2 \cdot (-3) \cdot 4i + (4i)^2 = 9 + 24i + 16i^2 \]

\[ 9 + 24i + 16(-1) = 9 + 24i - 16 = -7 + 24i \]

So, \((1 - 2i)^4 = -7 + 24i\).

Find the reciprocal of \((1 - 2i)^4\)

We need to find:

\[ \frac{1}{-7 + 24i} \]

To simplify this, multiply the numerator and denominator by the complex conjugate of the denominator:

\[ \frac{1}{-7 + 24i} \cdot \frac{-7 - 24i}{-7 - 24i} = \frac{-7 - 24i}{(-7)^2 - (24i)^2} \]

Simplify the denominator:

\[ (-7)^2 - (24i)^2 = 49 - 576i^2 = 49 - 576(-1) = 49 + 576 = 625 \]

So, the expression becomes:

\[ \frac{-7 - 24i}{625} = \frac{-7}{625} - \frac{24i}{625} \]

\[ \frac{1}{(1 - 2i)^4} = \frac{-7}{625} - \frac{24}{625}i \]

However, the problem asks for the form \( \frac{a}{b} \), so we can express the real and imaginary parts separately:

Real part: \( \frac{-7}{625} \), Imaginary part: \( \frac{-24}{625} \)

Thus, the final answer is:

\[ \frac{1}{(1 - 2i)^4} = \frac{-7 - 24i}{625} \]

Detailed Answer

a) To solve the equation \( 3(p + iq) = 2q - ip - 4(1 - i) \), we will equate the real and imaginary parts on both sides of the equation.

Expand both sides:

First, expand the left-hand side (LHS):

\[ 3(p + iq) = 3p + 3iq \]

Next, expand the right-hand side (RHS):

\[ 2q - ip - 4(1 - i) = 2q - ip - 4 + 4i \]

Equate the real and imaginary parts:

The equation \( 3p + 3iq = 2q - ip - 4 + 4i \) can be split into two separate equations by equating the real and imaginary parts:

1. Real part:

\[ 3p = 2q - 4 \]

2. Imaginary part:

\[ 3q = -p + 4 \]

We now have the following system of linear equations:

\[ (1) \ 3p = 2q - 4 \]

\[ (2) \ 3q = -p + 4 \]

Let's solve this system step by step.

From equation 2:

\[ 3q = -p + 4 \implies p = 4 - 3q \]

Substitute \( p = 4 - 3q \) into equation 1:

\[ 3(4 - 3q) = 2q - 4 \]

Expand and simplify:

\[ 12 - 9q = 2q - 4 \]

Bring all terms involving \( q \) to one side and constants to the other:

\[ 12 + 4 = 2q + 9q \]

\[ 16 = 11q \]

Solve for \( q \):

\[ q = \frac{16}{11} \]

Substitute \( q = \frac{16}{11} \) back into \( p = 4 - 3q \):

\[ p = 4 - 3\left(\frac{16}{11}\right) = 4 - \frac{48}{11} = \frac{44}{11} - \frac{48}{11} = -\frac{4}{11} \]

The solutions are:

\[ p = -\frac{4}{11}, \quad q = \frac{16}{11} \]

Detailed Answer

To find the other roots of the polynomial \( f(z) = 3z^4 - 12z^3 + 27z^2 - 30z + 18 \), given that \( z = 1 - i \) is a root, we can find the conjugate root.

Since the coefficients of the polynomial are real and \( z = 1 - i \) is a root, its complex conjugate \( z = 1 + i \) must also be a root.

The roots \( z = 1 - i \) and \( z = 1 + i \) correspond to the quadratic factor:

\[ (z - (1 - i))(z - (1 + i)) = (z - 1 + i)(z - 1 - i) \]

Expand this product:

\[ (z - 1)^2 - (i)^2 = z^2 - 2z + 1 + 1 = z^2 - 2z + 2 \]

Perform polynomial division:

Divide the original polynomial \( f(z) \) by the quadratic factor \( z^2 - 2z + 2 \) to find the remaining quadratic factor.

\[ f(z) = 3z^4 - 12z^3 + 27z^2 - 30z + 18 \]

Divide \( f(z) \) by \( z^2 - 2z + 2 \):

\[ 3z^4 - 12z^3 + 27z^2 - 30z + 18 = (z^2 - 2z + 2)(3z^2 - 6z + 9) \]

Solve the remaining quadratic equation:

Now, solve the quadratic equation \( 3z^2 - 6z + 9 = 0 \):

\[ z^2 - 2z + 3 = 0 \]

Use the quadratic formula to find the roots:

\[ z = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 3}}{2} = \frac{2 \pm \sqrt{4 - 12}}{2} = \frac{2 \pm \sqrt{-8}}{2} = \frac{2 \pm 2i\sqrt{2}}{2} = 1 \pm i\sqrt{2} \]

Detailed Answer

To find the values of \( p \), \( q \), and \( r \) for the polynomial \( Q(z) = z^3 + pz^2 + qz + r \), given that two of its roots are \(-1\) and \(-2 + 3i\), we can identify the conjugate root.

Since the coefficients of the polynomial are real and one of the roots is \(-2 + 3i\), its complex conjugate \(-2 - 3i\) must also be a root.

Express the polynomial in terms of its roots:

The polynomial \( Q(z) \) can be expressed as:

\[ Q(z) = (z - \alpha)(z - \beta)(z - \gamma) \]

where \( \alpha \), \( \beta \), and \( \gamma \) are the roots of the polynomial. Given the roots \(-1\), \(-2 + 3i\), and \(-2 - 3i\), we have:

\[ Q(z) = (z + 1)(z + 2 - 3i)(z + 2 + 3i) \]

Expand the polynomial:

First, multiply the two complex conjugate factors:

\[ (z + 2 - 3i)(z + 2 + 3i) = (z + 2)^2 - (3i)^2 = z^2 + 4z + 4 + 9 = z^2 + 4z + 13 \]

Now, multiply this result by the remaining factor:

\[ Q(z) = (z + 1)(z^2 + 4z + 13) \]

Expand the product:

\[ Q(z) = z(z^2 + 4z + 13) + 1(z^2 + 4z + 13) = z^3 + 4z^2 + 13z + z^2 + 4z + 13 \]

Combine like terms:

\[ Q(z) = z^3 + (4z^2 + z^2) + (13z + 4z) + 13 = z^3 + 5z^2 + 17z + 13 \]

Identify the coefficients:

Comparing \( Q(z) = z^3 + pz^2 + qz + r \) with the expanded form \( z^3 + 5z^2 + 17z + 13 \), we identify the coefficients:

\[ p = 5, \quad q = 17, \quad r = 13 \]

Detailed Answer

a) De Moivre's theorem states that for any integer \( n \):

\[ (\cos \phi + i \sin \phi)^n = \cos(n\phi) + i \sin(n\phi) \]

Applying this to \( w^4 \):

\[ w^4 = (\cos \phi + i \sin \phi)^4 = \cos(4\phi) + i \sin(4\phi) \]

b) Using the binomial theorem, expand \( w^4 = (\cos \phi + i \sin \phi)^4 \):

\[ (\cos \phi + i \sin \phi)^4 = \cos^4 \phi + 4i \cos^3 \phi \sin \phi - 6 \cos^2 \phi \sin^2 \phi - 4i \cos \phi \sin^3 \phi + \sin^4 \phi \]

Combine like terms:

\[ = (\cos^4 \phi - 6 \cos^2 \phi \sin^2 \phi + \sin^4 \phi) + i(4 \cos^3 \phi \sin \phi - 4 \cos \phi \sin^3 \phi) \]

c) From part (a), we have:

\[ w^4 = \cos(4\phi) + i \sin(4\phi) \]

From part (b), we have:

\[ w^4 = (\cos^4 \phi - 6 \cos^2 \phi \sin^2 \phi + \sin^4 \phi) + i(4 \cos^3 \phi \sin \phi - 4 \cos \phi \sin^3 \phi) \]

Equate the real and imaginary parts:

1. Real part:

\[ \cos(4\phi) = \cos^4 \phi - 6 \cos^2 \phi \sin^2 \phi + \sin^4 \phi \]

2. Imaginary part:

\[ \sin(4\phi) = 4 \cos^3 \phi \sin \phi - 4 \cos \phi \sin^3 \phi \]

Detailed Answer

a) Use de Moivre's theorem to show the given identities:

De Moivre's theorem states that for any integer \( n \):

\[ w^n = (\cos \phi + i \sin \phi)^n = \cos(n\phi) + i \sin(n\phi) \]

Similarly, for \( w^{-n} \):

\[ w^{-n} = (\cos \phi + i \sin \phi)^{-n} = \cos(n\phi) - i \sin(n\phi) \]

Now, compute \( w^n + w^{-n} \):

\[ w^n + w^{-n} = (\cos(n\phi) + i \sin(n\phi)) + (\cos(n\phi) - i \sin(n\phi)) = 2 \cos(n\phi) \]

Next, compute \( w^n - w^{-n} \):

\[ w^n - w^{-n} = (\cos(n\phi) + i \sin(n\phi)) - (\cos(n\phi) - i \sin(n\phi)) = 2i \sin(n\phi) \]

Thus, we have shown:

\[ w^n + w^{-n} = 2 \cos n \phi \]

\[ w^n - w^{-n} = 2i \sin n \phi \]

b) First, expand \( (w + w^{-1})^2 \):

\[ (w + w^{-1})^2 = w^2 + 2w w^{-1} + w^{-2} = w^2 + 2 + w^{-2} \]

Using the result from part (a):

\[ w^2 + w^{-2} = 2 \cos(2\phi) \]

Thus:

\[ (w + w^{-1})^2 = 2 \cos(2\phi) + 2 \]

But \( w + w^{-1} = 2 \cos \phi \), so:

\[ (2 \cos \phi)^2 = 2 \cos(2\phi) + 2 \]

\[ 4 \cos^2 \phi = 2 \cos(2\phi) + 2 \]

\[ 2 \cos^2 \phi = \cos(2\phi) + 1 \]

\[ \cos^2 \phi = \frac{1 + \cos(2\phi)}{2} \]

Next, expand \( (w - w^{-1})^2 \):

\[ (w - w^{-1})^2 = w^2 - 2w w^{-1} + w^{-2} = w^2 - 2 + w^{-2} \]

Using the result from part (a):

\[ w^2 - w^{-2} = 2i \sin(2\phi) \]

Thus:

\[ (w - w^{-1})^2 = 2i \sin(2\phi) - 2 \]

But \( w - w^{-1} = 2i \sin \phi \), so:

\[ (2i \sin \phi)^2 = 2i \sin(2\phi) - 2 \]

\[ -4 \sin^2 \phi = 2i \sin(2\phi) - 2 \]

Since the left side is real and the right side has an imaginary part, we equate the real parts:

\[ -4 \sin^2 \phi = -2 \]

\[ \sin^2 \phi = \frac{1 - \cos(2\phi)}{2} \]

c) First, expand \( (w + w^{-1})^3 \):

\[ (w + w^{-1})^3 = w^3 + 3w^2 w^{-1} + 3w w^{-2} + w^{-3} = w^3 + 3w + 3w^{-1} + w^{-3} \]

Using the result from part (a):

\[ w^3 + w^{-3} = 2 \cos(3\phi) \]

Thus:

\[ (w + w^{-1})^3 = 2 \cos(3\phi) + 3(w + w^{-1}) \]

But \( w + w^{-1} = 2 \cos \phi \), so:

\[ (2 \cos \phi)^3 = 2 \cos(3\phi) + 3(2 \cos \phi) \]

\[ 8 \cos^3 \phi = 2 \cos(3\phi) + 6 \cos \phi \]

\[ 4 \cos^3 \phi = \cos(3\phi) + 3 \cos \phi \]

\[ \cos^3 \phi = \frac{\cos(3\phi) + 3 \cos \phi}{4} \]

Next, expand \( (w - w^{-1})^3 \):

\[ (w - w^{-1})^3 = w^3 - 3w^2 w^{-1} + 3w w^{-2} - w^{-3} = w^3 - 3w + 3w^{-1} - w^{-3} \]

Using the result from part (a):

\[ w^3 - w^{-3} = 2i \sin(3\phi) \]

Thus:

\[ (w - w^{-1})^3 = 2i \sin(3\phi) - 3(w - w^{-1}) \]

But \( w - w^{-1} = 2i \sin \phi \), so:

\[ (2i \sin \phi)^3 = 2i \sin(3\phi) - 3(2i \sin \phi) \]

\[ -8i \sin^3 \phi = 2i \sin(3\phi) - 6i \sin \phi \]

\[ -8 \sin^3 \phi = 2 \sin(3\phi) - 6 \sin \phi \]

\[ \sin^3 \phi = \frac{3 \sin \phi - \sin(3\phi)}{4} \]