Complex Numbers and Quadratic Equations Questions and Answers

Answer: D

\(\overline{w} =\) \(-i\sqrt{3}\).

Conjugating \(z_{2}\) changes the sign of its imaginary part:

\[ \begin{aligned} \overline{z_{2}} &= -\tfrac{1}{2} + i\tfrac{\sqrt{3}}{2} \\ w &= z_{1} + \overline{z_{2}} = i\sqrt{3} \\ \overline{w} &= -i\sqrt{3} \end{aligned} \]

The real parts cancel, so \(w\) is purely imaginary. \(i\sqrt{3}\) is \(w\) itself, not its conjugate; the real options \(\pm\sqrt{3}\) and 1 would need the imaginary parts to cancel instead.

Answer: B

The solution set is \(\left[-2, \frac{7}{2}\right]\).

\(|u| \le 11\) means \(-11 \le u \le 11\):

\[ \begin{aligned} -11 \le 3 - 4x &\le 11 \\ -14 \le -4x &\le 8 \\ -2 \le x &\le \tfrac{7}{2} \end{aligned} \]

Dividing by \(-4\) reverses both inequalities. Check: \(x = -2\) gives \(|11| = 11\) and \(x = 3.5\) gives \(|-11| = 11\). \([-2, 7]\) forgets to divide 14 by 4, and the other intervals have the signs reversed.

Answer: B

\(x\) lies in \((11, 12]\).

Move everything to one side over the common denominator:

\[ \frac{2x - 21}{x - 11} - 3 = \frac{12 - x}{x - 11} \ge 0 \]

The fraction is non-negative when numerator and denominator have the same sign (or the numerator is 0): \(12 - x \ge 0\) with \(x - 11 \gt 0\) gives \(11 \lt x \le 12\); the other case is impossible. 12 itself is included (the fraction equals 0); 11 is not. Multiplying through by \(x - 11\) without checking its sign leads to the wrong intervals.

Answer: B

z lies on the real axis.

Put \(z = x + iy\) and \(\bar{z} = x - iy\), then separate real and imaginary parts.

\[ \begin{aligned} s(x + iy) + t(x - iy) + r &= 0 \\ (s + t)x + r + i(s - t)y &= 0 \end{aligned} \]

The imaginary part gives \((s - t)y = 0\). Since s and t are distinct, \(s - t \neq 0\), so y = 0 and z is real. (The real part then fixes \(x = -\frac{r}{s + t}\).) A point with y = 0 cannot lie on the imaginary axis or on the lines y = x, y = 2x or x + y = 0 unless it is the origin, which it is not because r is non-zero.

Answer: E

It is the circle with centre at (1,-1) and radius 5.

Write the condition as a distance: \(|z - z_{0}| = r\) is the circle of radius r about \(z_{0}\).

\[ |z - 1 + i| = |z - (1 - i)| = 5 \]

So the centre is the point \(1 - i\), that is (1, -1), and the radius is 5. Reading the centre as (-1, 1) flips both signs, and the options with radius \(\sqrt{5}\) or 25 confuse the radius with its square root or its square.

Answer: C

The value is |z| = 1.

Group the terms to factorise. Since \(-i \cdot i = 1\), the last two terms are \(-i(z + i)\).

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

So either z = -i, with |z| = 1, or \(z^{2} = i\), which gives \(|z|^{2} = |i| = 1\) and again |z| = 1. Every root lies on the unit circle, so 2, 3, 1/2 and 1/4 are impossible.

Answer: B

The real part is \(sin\frac{2\pi}{27}\).

Let \(\theta = \frac{2\pi}{27}\) and \(w = \sin\theta + i\cos\theta\). Then \(|w| = 1\), so \(\bar{w} = \frac{1}{w}\), and the numerator is \(1 + \bar{w}\).

\[ \begin{aligned} \frac{1 + \bar{w}}{1 + w} &= \frac{1 + \frac{1}{w}}{1 + w} = \frac{w + 1}{w(1 + w)} \\ &= \frac{1}{w} = \bar{w} = \sin\theta - i\cos\theta \end{aligned} \]

The real part is \(\sin\frac{2\pi}{27}\); the imaginary part is \(-\cos\frac{2\pi}{27}\), which is where the cosine option comes from. The options with 1 added keep a piece of the unsimplified numerator.

Answer: C

The solution set is \((-2, \infty)\).

Collect x on one side.

\[ \begin{aligned} 7x + 4 &\lt 9x + 8 \\ -4 &\lt 2x \\ x &\gt -2 \end{aligned} \]

The inequality is strict, so -2 itself is excluded and the bracket at -2 is round, which rules out \([-2, \infty)\). \((-\infty, -2)\) reverses the inequality, a slip that happens when dividing by a negative number that is not actually there.

Answer: C

The solution set is \(\left[\frac{29}{9}, \infty\right)\).

Multiply both sides by 35, which is positive, so the direction of the inequality is kept.

\[ \begin{aligned} 15(x + 3) &\le 42(x - 1) \\ 15x + 45 &\le 42x - 42 \\ 87 &\le 27x \Rightarrow x \ge \frac{29}{9} \end{aligned} \]

The inequality is not strict, so \(\frac{29}{9}\) itself is a solution and the bracket is square; \(\left(\frac{29}{9}, \infty\right)\) leaves it out. \(\left(-\infty, \frac{29}{9}\right)\) reverses the sign.

Answer: D

The points lie on a circle with radius 1.

For real \(\lambda\), \(\lambda + i\) and \(\lambda - i\) are complex conjugates, so they have the same modulus:

\[ \begin{aligned} \left|\frac{\lambda + i}{\lambda - i}\right| &= \frac{\sqrt{\lambda^{2} + 1}}{\sqrt{\lambda^{2} + 1}} \\ &= 1 \end{aligned} \]

Every point of A is at distance 1 from the origin, so A lies on the unit circle. The radii \(\sqrt{2}\), 2 and \(\frac{1}{2}\) do not arise, and the points are not on a line.