2. Use de Moivre's Theorem to solve each of these equations. Plot your solutions on the complex plane. \( \begin{array}{ll}\text { (i) } z^{2}=4 i & \text { (v) } z^{3}=8\end{array} \)

We wish to solve the equations using de Moivre’s Theorem and then plot the solutions on the complex plane. ───────────────────────────── Problem (i): Solve z² = 4i Step 1. Write the right‐hand side in polar form. The complex number 4i can be written as:   4i = 4 (cos(π/2) + i sin(π/2)) = 4 cis(π/2). Step 2. Let z = r cis(θ) and write the equation using de Moivre’s Theorem. Since   z² = (r cis(θ))² = r² cis(2θ), the equation becomes:   r² cis(2θ) = 4 cis(π/2). Step 3. Equate the moduli and the angles. a) Equate the moduli:   r² = 4 ⇒ r = 2  (since r ≥ 0) b) Equate the angles:   2θ = π/2 + 2πk  with k an integer. Thus:   θ = π/4 + πk     (for any integer k) Step 4. Write the solutions. Since z depends on k modulo 2 (because adding 2π in the exponent gives the same number), we get two distinct solutions:  For k = 0:   θ = π/4  ⇒  z = 2 cis(π/4) = 2 (cos(π/4) + i sin(π/4)) = 2 (√2/2 + i √2/2) = √2 + i √2.  For k = 1:   θ = π/4 + π = 5π/4  ⇒  z = 2 cis(5π/4) = 2 (cos(5π/4) + i sin(5π/4)) = 2 (–√2/2 – i √2/2) = –√2 – i √2. ───────────────────────────── Problem (v): Solve z³ = 8 Step 1. Write the right‐hand side in polar form. The number 8 is real and positive, so it can be written as:   8 = 8 cis(0). Step 2. Let z = r cis(θ) and apply de Moivre’s Theorem. We have   z³ = (r cis(θ))³ = r³ cis(3θ) = 8 cis(0). Step 3. Equate the moduli and the angles. a) Modulus equation:   r³ = 8 ⇒  r = 2. b) Angle equation:   3θ = 0 + 2πk  ⇒  θ = 2πk/3     (for any integer k) Step 4. Write the three distinct solutions. Since the cube roots repeat every 3 values, choose k = 0, 1, 2: For k = 0:   θ = 0  ⇒  z = 2 cis(0) = 2 (cos 0 + i sin 0) = 2. For k = 1:   θ = 2π/3  ⇒  z = 2 cis(2π/3) = 2 (cos(2π/3) + i sin(2π/3)) = 2 (–1/2 + i √3/2) = –1 + i√3. For k = 2:   θ = 4π/3  ⇒  z = 2 cis(4π/3) = 2 (cos(4π/3) + i sin(4π/3)) = 2 (–1/2 – i √3/2) = –1 – i√3. ───────────────────────────── Plotting the Solutions on the Complex Plane For (i) z² = 4i: • The first solution z = √2 + i √2 is located in the first quadrant, at the point (√2, √2). • The second solution z = –√2 – i √2 is located in the third quadrant, at the point (–√2, –√2). For (v) z³ = 8: • z = 2 lies on the positive real axis at (2, 0). • z = –1 + i√3 lies in the second quadrant at (–1, √3). • z = –1 – i√3 lies in the third quadrant at (–1, –√3). ───────────────────────────── Summary of Answers (i) z² = 4i has two solutions:   z = 2 cis(π/4) = √2 + i√2   and  z = 2 cis(5π/4) = –√2 – i√2. (v) z³ = 8 has three solutions:   z = 2 cis(0) = 2,   z = 2 cis(2π/3) = –1 + i√3,   z = 2 cis(4π/3) = –1 – i√3. You can now plot these points on the complex plane to see their geometric distribution.