Complex Numbers and Quadratic Equations Questions and Answers

Answer: C

\(x\) lies in \([-35, -20) \cup (0, 15]\).

\(10 \lt |x + 10| \leq 25\) means \(x + 10\) is more than 10 but at most 25 away from 0, on either side:

\[ \begin{aligned} 10 \lt x + 10 \leq 25 &\Rightarrow 0 \lt x \leq 15 \\ -25 \leq x + 10 \lt -10 &\Rightarrow -35 \leq x \lt -20 \end{aligned} \]

The strict inequality at 10 makes 0 and -20 excluded (round brackets), and the non-strict one at 25 makes 15 and -35 included (square brackets). The other options have a wrong endpoint.

Answer: B

\(|z|\) is 2.

Square both sides, using \(|z + a|^{2} = |z|^{2} + 2a\,\mathrm{Re}(z) + a^{2}\) for real \(a\):

\[ \begin{aligned} |z|^{2} + 8x + 16 &= 4(|z|^{2} + 2x + 1) \\ |z|^{2} + 8x + 16 &= 4|z|^{2} + 8x + 4 \\ 3|z|^{2} &= 12 \\ |z| &= 2 \end{aligned} \]

where \(x = \mathrm{Re}(z)\). The \(8x\) terms cancel, so every \(z\) satisfying the condition lies on the circle \(|z| = 2\). 4 is \(|z|^{2}\), not \(|z|\).

Answer: A

\(z^{2} = \frac{i}{2}\).

First find \(z\) by multiplying numerator and denominator by the conjugate \(3 + i\):

\[ \begin{aligned} z &= \frac{2 + i}{3 - i} \cdot \frac{3 + i}{3 + i} \\ &= \frac{6 + 5i + i^{2}}{10} = \frac{5 + 5i}{10} = \frac{1 + i}{2} \end{aligned} \]

\[ \begin{aligned} z^{2} &= \frac{(1 + i)^{2}}{4} = \frac{1 + 2i - 1}{4} \\ &= \frac{2i}{4} = \frac{i}{2} \end{aligned} \]

\(\frac{1 + i}{2}\) is \(z\) itself, not \(z^{2}\), and \(\frac{-i}{2}\) comes from a sign slip in \(i^{2} = -1\).

Answer: D

The imaginary part is \(\frac{-\sqrt{3}}{2}\).

Multiply numerator and denominator by the conjugate of the denominator, \(1 - i\sqrt{3}\):

\[ \begin{aligned} \frac{(1 - i\sqrt{3})^{2}}{1 + 3} &= \frac{1 - 2i\sqrt{3} - 3}{4} \\ &= \frac{-2 - 2i\sqrt{3}}{4} \\ &= -\frac{1}{2} - \frac{\sqrt{3}}{2}i \end{aligned} \]

The imaginary part is the coefficient of \(i\), \(-\frac{\sqrt{3}}{2}\). \(-\frac{1}{2}\) is the real part, and \(\frac{\sqrt{3}}{2}\) has the wrong sign.

Answer: E

The sum is -1.

The terms are \(i^{2}, i^{4}, i^{6}, \ldots\), and since \(i^{2} = -1\) and \(i^{4} = 1\), they alternate: \(-1, 1, -1, 1, \ldots\)

Each pair \(-1 + 1\) adds to 0. Twenty-five terms make 12 pairs (summing to 0) plus one more term, the 25th, which is \(i^{50} = i^{2} = -1\).

So the sum is \(-1\). 0 would be right for an even number of terms, and \(\pm i\) cannot arise because every term is an even power of \(i\), which is real.

Answer: C

\(x\) belongs to \((-\infty, 1) \cup (2, 3)\).

Bring everything to one side and factor:

\[ \begin{aligned} (x - 1)(x^{2} - 5x + 7) - (x - 1) &\lt 0 \\ (x - 1)(x^{2} - 5x + 6) &\lt 0 \\ (x - 1)(x - 2)(x - 3) &\lt 0 \end{aligned} \]

The critical points are 1, 2 and 3. The product is negative for \(x \lt 1\) and for \(2 \lt x \lt 3\), and positive in between and beyond 3. The inequality is strict, so the endpoints are excluded.

Options with \(-1\) misplace the first root, and those with square brackets wrongly include 2 or 3, where the product is zero.

Answer: E

The solution set is \(\mathbb{R} - \{-1, 0, 1\}\).

For \(x \gt 0\), the AM-GM inequality gives \(x + \frac{1}{x} \ge 2\), with equality only at \(x = 1\). For \(x \lt 0\) the expression is the negative of the same thing, so \(\left|x + \frac{1}{x}\right| \ge 2\) for every \(x \ne 0\), with equality only at \(x = \pm 1\).

So the strict inequality \(\gt 2\) holds for all real \(x\) except \(-1\) and \(1\), and \(x = 0\) is excluded because \(\frac{1}{x}\) is undefined there.

\(\mathbb{R}\) and \(\mathbb{R} - \{0\}\) forget the points of equality, and the other options remove only some of the three excluded values.

Answer: A

The equation represents the line \(y = x + 1\).

Write \(z = x + iy\) and multiply out.

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

The imaginary part is \(y - x\), so \(y - x = 1\), that is \(y = x + 1\). Taking the real part by mistake gives \(x + y = 1\), the option \(y = 1 - x\), and a sign slip in \(-i^{2}y\) produces the other lines.

Answer: D

The value is 0.

Write the numbers in polar form: \(\frac{\sqrt{3}}{2} \pm \frac{i}{2} = \cos 30^\circ \pm i\sin 30^\circ\). By De Moivre's theorem, cubing triples the angle.

\[ \begin{aligned} (\cos 30^\circ + i\sin 30^\circ)^{3} &= \cos 90^\circ + i\sin 90^\circ = i \\ (\cos 30^\circ - i\sin 30^\circ)^{3} &= \cos 90^\circ - i\sin 90^\circ = -i \end{aligned} \]

The sum is \(i + (-i) = 0\). The two terms are conjugates, so their sum is twice the real part of either, \(2\cos 90^\circ = 0\). \(2\sqrt{3}\) and \(3\sqrt{3}\) come from expanding the cubes with a sign error.

Answer: B

x lies in [-2, 2].

Solve each inequality and take the common part.

\[ \begin{aligned} \frac{21x - 6}{4} \le 9 &\Rightarrow 21x \le 42 \Rightarrow x \le 2 \\ \frac{x - 1}{3} \ge -1 &\Rightarrow x - 1 \ge -3 \Rightarrow x \ge -2 \end{aligned} \]

Both conditions hold for \(-2 \le x \le 2\). Both ends are included because the inequalities are not strict. [-2, 4] and [-2, 1] come from slips in clearing the fractions.