Complex Numbers and Quadratic Equations Questions and Answers

Answer: B

The sum equals \(i - 1\).

Take out the common factor \((1 + i)\). The powers of i repeat with period 4 and each block of four sums to 0 (\(i - 1 - i + 1 = 0\)). Since \(2025 = 4 \times 506 + 1\), only the last term survives:

\[ \begin{aligned} \sum_{n=1}^{2025} i^{n} &= i^{2025} = i \\ i(1 + i) &= i + i^{2} = i - 1 \end{aligned} \]

\(i + 1\) forgets that \(i^{2} = -1\), and the other options come from leaving an extra term or a sign in the cycle.

The trick in all such sums is that \(i + i^{2} + i^{3} + i^{4} = 0\), so only the remainder of the number of terms on division by 4 matters.

Answer: D

\(x - y\) equals −91.

Expand by the binomial theorem:

\[ \begin{aligned} (6 + i)^{3} &= 216 + 3(36)i + 3(6)i^{2} + i^{3} \\ &= 216 + 108i - 18 - i = 198 + 107i \end{aligned} \]

So \(x + iy = -198 - 107i\), giving \(x = -198\), \(y = -107\) and \(x - y = -198 + 107 = -91\).

91 drops the overall minus sign, −107 is just y, and ±93 comes from a slip in the middle terms.

Answer: E

The condition gives \(y = x\).

\(|z - i| = |z - 1|\) says z is equidistant from the points i = (0, 1) and 1 = (1, 0), so it lies on the perpendicular bisector of the segment joining them. Algebraically:

\[ \begin{aligned} x^{2} + (y - 1)^{2} &= (x - 1)^{2} + y^{2} \\ -2y + 1 &= -2x + 1 \Rightarrow y = x \end{aligned} \]

\(y = -x\) is the line through the two points themselves turned the wrong way, and the shifted lines do not pass through the midpoint \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

Answer: A

a, b, c are in G.P. and \(a + b + c = 6\).

A sum of squares of real numbers is zero only if each square is zero, so \(a = b = c = 2\). Three equal non-zero numbers form a G.P. with common ratio 1 (and also an A.P.), and their sum is 6.

The options with sums 4, 8 or 16 have the wrong total, and 'not in G.P.' is false because a constant non-zero sequence is geometric.

Answer: D

The solution set is the empty set.

Dividing by −12 reverses the inequality: \(x \lt -\frac{38}{12} \approx -3.17\). Natural numbers are positive, so none satisfies it.

{1, 2, 3}, {1, 2} and {1} come from forgetting to reverse the inequality, and {−1, −2, −3} are not natural numbers (and do not satisfy \(x \lt -3.17\) anyway).

Answer: A

The solution set is \(\left(-\infty, \frac{156}{19}\right)\).

Combine the terms with denominator 12:

\[ \begin{aligned} x\left(1 + \frac{1}{4} + \frac{1}{3}\right) &= \frac{19x}{12} \lt 13 \\ x &\lt \frac{156}{19} \end{aligned} \]

The intervals that extend to \(+\infty\) reverse the inequality, and 154 or 17 come from arithmetic slips.

Answer: D

\(|z|\) equals \(-\sec\theta\).

For \(z = 1 + i\tan\theta\):

\[ \begin{aligned} |z| &= \sqrt{1 + \tan^{2}\theta} \\ &= \sqrt{\sec^{2}\theta} = |\sec\theta| \end{aligned} \]

For \(\pi \lt \theta \lt \frac{3\pi}{2}\), \(\theta\) is in the third quadrant, where \(\cos\theta\) and so \(\sec\theta\) are negative. A modulus must be positive, so \(|\sec\theta| = -\sec\theta\).

The trap is \(\sec\theta\), which would be right in the first or fourth quadrant but is negative here. 1 + tan θ and 2tan θ add the parts instead of combining them as a square root.

Answer: C

\(z^{-1}\) is \(\frac{1-i}{2}\).

Simplify z by multiplying numerator and denominator by the conjugate \(2 + i\):

\[ \begin{aligned} z &= \frac{(3+i)(2+i)}{(2-i)(2+i)} = \frac{6 + 5i + i^{2}}{5} \\ &= \frac{5 + 5i}{5} = 1 + i \end{aligned} \]

\[ \begin{aligned} z^{-1} &= \frac{1}{1+i} = \frac{1-i}{(1+i)(1-i)} \\ &= \frac{1-i}{2} \end{aligned} \]

1 + i is z itself, and \(1 - i\) is its conjugate, which equals \(z^{-1}\) only when \(|z| = 1\); here \(|z|^{2} = 2\), so the conjugate must be divided by 2.

Answer: E

The minimum value is 2.

\(|z - 2|\) and \(|z - 4|\) are the distances of the point z from the points 2 and 4 on the real axis. By the triangle inequality,

\[ |z-2| + |z-4| \ge |(z-2) - (z-4)| = 2 \]

Equality holds for every z on the segment joining 2 and 4, for example z = 3, which gives 1 + 1 = 2. The sum can never be smaller than the distance between the two fixed points, so 2 is the minimum. The values 4, 6, \(\sqrt{6}\) and \(\sqrt{20}\) are all attained at points off that segment, so none is the least.

Answer: B

The new position is 1.

Write z in polar form: \(\frac{1}{\sqrt{2}}(1 + i) = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = e^{i\pi/4}\), a point at distance 1 from the origin at angle \(\frac{\pi}{4}\).

A clockwise rotation through \(\frac{\pi}{4}\) multiplies by \(e^{-i\pi/4}\):

\[ e^{i\pi/4} \cdot e^{-i\pi/4} = e^{0} = 1 \]

The point lands on the positive real axis at 1. \(\frac{1}{\sqrt{2}}(1 - i)\) is the result of rotating clockwise through \(\frac{\pi}{2}\) instead, and 2 and \(1 - i\) change the distance from the origin, which a rotation cannot do.