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Problem 1: For this first problem, I probably did not solve it the best way, but I did it in a way that would let me follow the example in the book. I hope my work is clear. z = 27cis(240°) z = 27cis(4pi/3) I converted the degrees to radians. z = 27(cos(4pi/3) + i sin(4pi/3)) Now it is in a form that makes it easy to follow the examples in the book. z^1/3 = 27^1/3[cos((4pi/3)/3 + 4kpi/3) + i sin(4pi/9 + 4kpi/3)] it is important to note that k can be equal to 0, 1, and 2. So, we now need to insert each of these values into the equation.

When k is equal to 0, we have z^1/3 = 3(cos(4pi/9) + i sin(4pi/9)), 4kpi/3 cancels out. I will now convert 4pi/9 into degrees. 4pi/9 * 180/pi = 4*20*9/9 = 80° z^1/3 = 3(cos(80°) + i sin(80°)) Now from here I am asked to cube my answer. I am not sure if this is correct, but here is what I came up with. z = 27(cis(240°)). Again, not sure if this is correct.

When k is 1, we have z^1/3 = 3[cos(4pi/9 + 12pi/9) + I sin(4pi/9 + 12pi/9)] z^1/3 = 3(cos(16pi/9) + I sin(16pi/9)) We will now convert the radians into degrees. 16pi/9 * 180/pi = 16 * 20 * 9/9 = 320° Therefore, z^1/3 = 3(cos(320°) + i sin(320°)) Cubing it, we get z = 27(cis(960°)), or z = 27(cis(240°)) after subtracting 360° twice. Hope this is right, I am really not sure.

When k is 2, we have z^1/3 = 3(cos(4pi/9 + 24pi/9) + i sin(4pi + 24pi/9)) z^1/3 = 3(cos(28pi/9) + i sin(28pi/9))

converting degrees into radains, we get the following. 28pi/9 * 180/pi = 28 * 20 * 9/9 = 560° - 360° = 200°. Thus, z^1/3 = 3(200°) + i sin(200°)) Now we cube it. z = 27(600°) ´+ i sin(600°). We then subtract 360° from 600° z = 27(cis(240°))

Problem 2: Evaluate [5sqrt(3)(sqrt(3)/2 +i/2)]^10 For this problem, we are going to need the following formula: 2^n = r^n[cos(n ϴ) + i sin(nϴ)] We can see that r = 5sqrt(3), n = 10, and ϴ = 30° Now these values can be inserted into the formula. z^10 = 5sqrt(3)^10[cos(10*30°) + i sin(10 * 30°)] z^10 = 9[cos(300°) + i sin(300°) 9[1/2 + i(-sqrt(3)/2)] = 9/2 – 9sqrt(3)/2 i Problem 3: Find z1/z2 in polar form z1 = 21cis(135°) z2 = 3cis(75°) To solve this, we will make use of the formula z = z1/z2 = r1/r2 cis(ϴ1 - ϴ2) = 21/3 cis(135° - 75°) = 7cis(60°) = 7(1/2 + i sqrt3/2) z = 7/2 + 7sqrt(3)/2 i...