Complex Numbers and Quadratic Equations Questions and Answers

Answer: E

\(z = 4 - 5i\).

Let \(z = a + ib\), so \(Re(z) = a\) and \(Im(z) = b\). Rationalise each fraction; all three denominators have modulus squared 5:

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

\[ \begin{aligned} 2a + b &= 3 \\ a + 2b &= -6 \\ \Rightarrow a = 4,\ b &= -5 \end{aligned} \]

Check: \(2(4) - 5 = 3\) and \(4 - 10 = -6\). The options \(5 - 4i\) and \(6 - 5i\) fail the first equation, and \(-4 + 5i\) has both signs reversed.

Answer: B

The value is 4096.

Square the bracket first:

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

\[ \begin{aligned} (2 + 2i)^{2} &= 4 + 8i + 4i^{2} = 8i \\ \left(\alpha + \tfrac{i}{\alpha}\right)^{16} &= (2 + 2i)^{8} = (8i)^{4} \\ &= 4096\,i^{4} = 4096 \end{aligned} \]

The option 2048 is half the answer, and 4048 and 5096 are look-alike numbers; \(8^{4} = 4096\) is the only power that fits.

Answer: A

The solution set is \(\left(\frac{7}{3}, \infty\right)\).

\[ \begin{aligned} 6(2x + 3) + x &\gt 53 - 2x \\ 12x + 18 + x &\gt 53 - 2x \\ 15x &\gt 35 \\ x &\gt \tfrac{7}{3} \end{aligned} \]

Dividing by the positive number 15 does not reverse the inequality, so every \(x\) greater than \(\frac{7}{3}\) works. The set \(\left(-\infty, \frac{7}{3}\right)\) reverses the sign, and the bounded intervals add a limit the inequality does not have.

Answer: E

The range is \(3 \lt a \lt 5\).

With right angles at A and D, the parallel sides are AB and CD and AD is the height. Drop a perpendicular from C to AB: it has length \(3a\) and cuts off \(AB - CD = 4a\).

\[ \begin{aligned} BC &= \sqrt{(3a)^{2} + (4a)^{2}} = 5a \\ \text{perimeter} &= (7a + 1) + 5a + (3a + 1) + 3a \\ &= 18a + 2 \end{aligned} \]

\[ \begin{aligned} 56 \lt 18a + 2 &\lt 92 \\ 54 \lt 18a &\lt 90 \\ 3 \lt a &\lt 5 \end{aligned} \]

Check the end points: \(a = 3\) gives a perimeter of exactly 56 and \(a = 5\) exactly 92, both excluded by the strict inequalities.

Answer: A

The value of a is 1.

Multiply numerator and denominator by the conjugate of \(i - 2\), which is \(-i - 2\):

\[ \begin{aligned} \left(a - \tfrac{i}{2}\right)(-2 - i) &= -2a - ai + i + \tfrac{i^{2}}{2} \\ &= \left(-2a - \tfrac{1}{2}\right) + (1 - a)i \end{aligned} \]

The denominator becomes \((i - 2)(-i - 2) = 5\), a real number, so \(\text{Im}(z) = \frac{1 - a}{5}\).

Setting this to 0 gives a = 1. Check: with a = 1, z = \(\frac{1 - i/2}{i - 2} = -\frac{1}{2}\), which is real.

Answer: B

The area is \(9\sqrt{3}\) square units.

The two vertices are \(-3 - 3i\) and \(3 - 3i\), the points (-3, -3) and (3, -3). They lie on the same horizontal line, so the side is 6.

\[ \begin{aligned} \text{Area} &= \frac{\sqrt{3}}{4}a^{2} \\ &= \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \end{aligned} \]

18 is \(\frac{1}{2} \times 6 \times 6\), the area of a right isosceles triangle with legs 6, not of an equilateral triangle of side 6. The height of the equilateral triangle is \(3\sqrt{3}\), and \(\frac{1}{2} \times 6 \times 3\sqrt{3} = 9\sqrt{3}\) confirms the answer.

Answer: E

The value is \(\frac{-1}{2^{2026}}\).

The denominator is a product of conjugates: \((3 + i)(3 - i) = 9 + 1 = 10\). So the bracket is \(\frac{5i}{10} = \frac{i}{2}\).

\[ \left(\frac{i}{2}\right)^{2026} = \frac{i^{2026}}{2^{2026}} \]

Powers of i repeat every 4, and 2026 = 4 × 506 + 2, so \(i^{2026} = i^{2} = -1\). The answer is \(-\frac{1}{2^{2026}}\).

Forgetting the sign gives \(\frac{1}{2^{2026}}\); halving the exponent to 1013 confuses \(i^{2}\) with the power of 2.

Answer: A

|z| = 5.

Take the modulus of both sides. Since |z| is a real number, \(|z\,|z|| = |z| \cdot |z| = |z|^{2}\).

\[ \begin{aligned} |z|^{2} &= |24 + 7i| = \sqrt{576 + 49} \\ &= \sqrt{625} = 25 \\ |z| &= 5 \end{aligned} \]

25 is \(|z|^{2}\), the trap of stopping one step early. Check: z = \(\frac{24 + 7i}{5}\) has modulus 5, and 5z = 24 + 7i.

Answer: C

The solution set is \((-\infty, 18)\).

\[ \begin{aligned} 8 + 3x &\gt 4x - 12 + 2 \\ 8 + 3x &\gt 4x - 10 \\ 18 &\gt x \end{aligned} \]

So every x less than 18 works. \((18, \infty)\) reverses the inequality, a slip that happens when x is collected on the left and the sign of its coefficient is not handled. Check: x = 0 gives 8 > -10, true.

Answer: D

The sum is \(i - 1\).

Powers of \(i\) repeat in blocks of four, \(i, -1, -i, 1\), and each block of four consecutive powers adds to 0. From \(i^{13}\) to \(i^{2026}\) there are \(2026 - 13 + 1 = 2014\) terms, and \(2014 = 4 \times 503 + 2\).

So 503 complete blocks contribute 0, leaving the last two terms:

\[ \begin{aligned} i^{2025} + i^{2026} &= i^{4 \cdot 506 + 1} + i^{4 \cdot 506 + 2} \\ &= i + (-1) = i - 1 \end{aligned} \]

Equivalently, the first two terms \(i^{13} = i\) and \(i^{14} = -1\) are the ones left over once the complete blocks cancel. The other options use the wrong remainder or sign.