answer:To solve this problem, we analyze each option step by step: **Option A:** - The complex number z_{1}=2-2i corresponds to point P_{1} in the complex plane. By definition, the real part of z_{1} gives the x-coordinate, and the imaginary part (without the i) gives the y-coordinate. Therefore, P_{1} has coordinates (2, -2). - Thus, boxed{text{A is correct}}. **Option B:** - The conjugate of a complex number z=a+bi is given by bar{z}=a-bi. For z_{1}=2-2i, the conjugate is bar{z_{1}}=2-(-2i)=2+2i. - Therefore, boxed{text{B is correct}}. **Options C and D:** - Let's denote z_{2}=x+yi where x,yin mathbb{R}. The condition |z_{2}-i|=1 translates to |x+yi-i|=|x+(y-1)i|=sqrt{x^2+(y-1)^2}=1. - This implies x^{2}+(y-1)^{2}=1, which is the equation of a circle centered at (0,1) with radius 1. - The distance between any point z_{2} on this circle and z_{1} is given by |z_{2}-z_{1}|=|x+yi-(2-2i)|=sqrt{(x-2)^2+(y+2)^2}. - To find the maximum and minimum values of this distance, we consider the geometric interpretation. The maximum distance occurs when z_{2} is the point on the circle farthest from P_{1}(2,-2), and the minimum distance occurs when z_{2} is the closest point on the circle to P_{1}. - The distance from the center of the circle (0,1) to P_{1}(2,-2) is sqrt{(2-0)^2+(-2-1)^2}=sqrt{4+9}=sqrt{13}. - Since the radius of the circle is 1, the maximum distance is sqrt{13}+1 and the minimum distance is sqrt{13}-1. - Therefore, boxed{text{C is correct}} and boxed{text{D is incorrect}}. Given the analysis, the correct options are boxed{text{A, B, and C}}.

answer:Given: 1. Mistaken calculation: a - (b - 2c) = 19 2. Correct calculation: a - b - 2c = 7 Using parentheses priority: - Bill's mistake turns the expression into a - b + 2c = 19. Equating the correct and mistaken equations: [ a - b - 2c = 7 ] [ a - b + 2c = 19 ] Subtract the first equation from the second: [ (a - b + 2c) - (a - b - 2c) = 19 - 7 ] [ 4c = 12 ] [ c = 3 ] Substitute c = 3 into either equation: [ a - b - 2 times 3 = 7 ] [ a - b - 6 = 7 ] [ a - b = 13 ] Conclusion: The value of a - b calculated correctly equates to boxed{13}.

answer:Let s be the common ratio. Then b_2 = 2s and b_3 = 2s^2. Thus, the expression to minimize is: [3b_2 + 4b_3 = 3(2s) + 4(2s^2) = 6s + 8s^2.] We can rewrite this quadratic expression as: [8s^2 + 6s = 8left(s + frac{3}{16}right)^2 - frac{9}{8}.] The minimum value of this expression occurs when the squared term is zero, which is when s = -frac{3}{16}. Plugging this value back into the expression gives: [8left(-frac{3}{16} + frac{3}{16}right)^2 - frac{9}{8} = -frac{9}{8}.] Thus, the minimum value of 3b_2 + 4b_3 is boxed{-frac{9}{8}}.

answer:1. We begin by considering the left-hand side of the given inequality: sum_{j=2}^{n^{k}} frac{1}{j} 2. We can decompose the above sum into smaller sums: sum_{j=2}^{n^{k}} frac{1}{j} = left( sum_{j=2}^{n} frac{1}{j} right) + left( sum_{j=n+1}^{n^2} frac{1}{j} right) + ldots + left( sum_{j=n^{k-1}+1}^{n^k} frac{1}{j} right) 3. This can be expressed as a sum of sums: sum_{j=2}^{n^{k}} frac{1}{j} = sum_{i=1}^{k} left( sum_{j=n^{i-1}+1}^{n^i} frac{1}{j} right) 4. For any ( i in mathbb{N} ), we consider the inner sum: sum_{j=n^{i-1}+1}^{n^i} frac{1}{j} 5. We can partition this sum to illustrate the inequality: sum_{j=n^{i-1}+1}^{n^i} frac{1}{j} = left( frac{1}{n^{i-1}+1} + cdots + frac{1}{2n^{i-1}} right) + left( frac{1}{2n^{i-1}+1} + cdots + frac{1}{3n^{i-1}} right) + ldots + left( frac{1}{(n-1)n^{i-1}+1} + cdots + frac{1}{n cdot n^{i-1}} right) 6. Summarizing, sum_{j=n^{i-1}+1}^{n^i} frac{1}{j} = sum_{m=2}^{n} left( frac{1}{(m-1)n^{i-1}+1} + cdots + frac{1}{m n^{i-1}} right) 7. Notice that: sum_{m=2}^{n} left( frac{1}{(m-1)n^{i-1}+1} + cdots + frac{1}{m n^{i-1}} right) > sum_{m=2}^{n} left( n^{i-1} frac{1}{m n^{i-1}} right) = sum_{m=2}^{n} frac{1}{m} 8. Therefore: sum_{j=n^{i-1}+1}^{n^i} frac{1}{j} > sum_{m=2}^{n} frac{1}{m} 9. Summing the inequalities over ( i ) from 1 to ( k ): sum_{i=1}^{k} left( sum_{j=n^{i-1}+1}^{n^i} frac{1}{j} right) > sum_{i=1}^{k} left( sum_{m=2}^{n} frac{1}{m} right) 10. Which simplifies to: sum_{j=2}^{n^k} frac{1}{j} > k sum_{j=2}^{n} frac{1}{j} [ boxed{} ]