This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Okay, let's solve question 8 from the image you just sent. --- Question 8 i) a): Given that z_1 = sqrt(3)+i and z_2 = 1-3i, find (z_1)/(z_2) in the form a+bi, where a and b are real. Step 1: Write down the given complex numbers. z_1 = sqrt(3)+i z_2 = 1-3i Step 2: Set up the division. (z_1)/(z_2) = sqrt(3)+i1-3i Step 3: Multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of 1-3i is 1+3i. (z_1)/(z_2) = sqrt(3)+i1-3i × (1+3i)/(1+3i) Step 4: Multiply the numerators. (sqrt(3)+i)(1+3i) = sqrt(3)(1) + sqrt(3)(3i) + i(1) + i(3i) = sqrt(3) + 3sqrt(3)i + i + 3i^2 Substitute i^2 = -1: = sqrt(3) + 3sqrt(3)i + i - 3 = (sqrt(3)-3) + (3sqrt(3)+1)i Step 5: Multiply the denominators. (1-3i)(1+3i) = 1^2 - (3i)^2 = 1 - 9i^2 Substitute i^2 = -1: = 1 - 9(-1) = 1 + 9 = 10 Step 6: Combine the simplified numerator and denominator and express in the form a+bi. (z_1)/(z_2) = (sqrt(3)-3) + (3sqrt(3)+1)i10 (z_1)/(z_2) = sqrt(3)-310 + 3sqrt(3)+110i The quotient (z_1)/(z_2) is sqrt(3)-310 + 3sqrt(3)+110i. --- Question 8 i) b): Given that z_1 = sqrt(3)+i and z_2 = 1-3i, find z_1^5 and |z_2^6|. Part 1: Find z_1^5 Step 1: Convert z_1 to polar form r_1(_1 + i_1). z_1 = sqrt(3)+i The modulus r_1 = |z_1| = sqrt((3))^2 + 1^2 = sqrt(3+1) = sqrt(4) = 2. The argument _1 = z_1. Since z_1 is in the first quadrant: _1 = (1)/(sqrt(3)) _1 = ()/(6) So, z_1 = 2((()/(6)) + i(()/(6))). Step 2: Use De Moivre's Theorem to find z_1^5. For z_1^n = r_1^n((n_1) + i(n_1)), the argument is n_1. z_1^5 = 5_1 = 5 × ()/(6) = (5)/(6) The argument z_1^5 is (5)/(6). Part 2: Find |z_2^6| Step 1: Find the modulus of z_2. z_2 = 1-3i |z_2| = sqrt(1^2 + (-3)^2) = sqrt(1+9) = sqrt(10) Step 2: Calculate |z_2^6|. Using the property |z^n| = |z|^n: |z_2^6| = (|z_2|)^6 = (sqrt(10))^6 (sqrt(10))^6 = (10^1/2)^6 = 10^(1/2) × 6 = 10^3 = 1000 The modulus |z_2^6| is 1000. --- Question 8 ii): Show that the roots of the equation z^3 - 1 = 0 lies at the vertices of an equilateral triangle. Step 1: Find the roots of the equation z^3 - 1 = 0. z^3 = 1 We need to find the cube roots of 1. In polar form, 1 = 1((0) + i(0)). The n-th roots of a complex number r( + i) are given by: z_k = [n]r((( + 2k)/(n)) + i(( + 2k)/(n))) For z^3=1, we have r=1, =0, and n=3. The values for k are 0, 1, 2. For k=0: z_0 = [3]1((0 + 2(0