Complex Numbers and Quadratic Equations Questions and Answers

Answer: D

The sum is 3.

For any complex number, \(z\overline{z} = |z|^{2}\), and the modulus of a quotient is the quotient of the moduli.

\[ \begin{aligned} |z_1|^{2} &= \frac{5^{2} + 7^{2}}{7^{2} + 5^{2}} = \frac{74}{74} = 1 \\ |z_2|^{2} &= \frac{3^{2} + 2^{2}}{3^{2} + 2^{2}} = 1 \\ |z_3|^{2} &= \frac{1 + 121}{121 + 1} = 1 \end{aligned} \]

So the sum is \(1 + 1 + 1 = 3\). Since \(z\overline{z}\) is always real, the options \(1 + 2i\) and \(1 - 2i\) are impossible at once.

Answer: C

The value is \(-4i^{n}\).

Split off four factors of \(1 - i\):

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

Hence the expression equals \(i^{n} \cdot (-4) = -4i^{n}\). The option \(4i^{n}\) misses the sign from \((-2i)^{2} = -4\), and \(\frac{i^{n}}{4}\) inverts the factor.

Answer: D

The solution is \(-14 \lt x \lt -4\) or \(6 \lt x \lt 16\).

\(5 \lt |x - 1| \lt 15\) means \(x - 1\) lies between 5 and 15 or between -15 and -5:

\[ \begin{aligned} 5 \lt x - 1 \lt 15 &\Rightarrow 6 \lt x \lt 16 \\ -15 \lt x - 1 \lt -5 &\Rightarrow -14 \lt x \lt -4 \end{aligned} \]

Check with an endpoint: \(x = 16\) gives \(|15| = 15\), just outside, and \(x = -4\) gives \(|-5| = 5\), just outside. Every other option has at least one wrong bound, for instance \(-3\) or 17.

Answer: B

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

Bring everything to one side; do not multiply by \(x - 2\), whose sign is unknown:

\[ \begin{aligned} \frac{x - 3}{x - 2} - 1 &\geq 0 \\ \frac{x - 3 - (x - 2)}{x - 2} &= \frac{-1}{x - 2} \geq 0 \end{aligned} \]

A fraction with numerator -1 is non-negative only when the denominator is negative, so \(x - 2 \lt 0\), that is \(x \lt 2\).

\((-\infty, 3)\) wrongly includes values such as \(x = 2.5\), where the fraction is \(-1\). \([0, \infty)\), \((-9, \infty)\) and \((0, 8)\) all contain numbers above 2.

Answer: E

The product equals \(-i\).

Multiplying powers of \(i\) adds the exponents:

\[ \begin{aligned} 1 + 2 + \dots + 2026 &= \frac{2026 \times 2027}{2} \\ &= 2053351 \end{aligned} \]

Powers of \(i\) repeat every 4, so only the remainder on division by 4 matters: \(2053351 = 4 \times 513337 + 3\). Hence the product is \(i^{3} = -i\).

A remainder of 1, 2 or 0 would have given \(i\), \(-1\) or 1, which is how the other options arise.

Answer: B

\(z_{1}\) is a purely real number and \(z_{2}\) is a purely imaginary number.

A number plus its conjugate is real, and a number minus its conjugate is purely imaginary. Since \((7 - i\sqrt{5})^{2}\) is the conjugate of \((7 + i\sqrt{5})^{2}\), and \((3 - 2i)^{3}\) is the conjugate of \((3 + 2i)^{3}\):

\[ \begin{aligned} z_{1} &= 2\,\text{Re}(7 + i\sqrt{5})^{2} = 2(49 - 5) = 88 \\ z_{2} &= 2i\,\text{Im}(3 + 2i)^{3} = 92i \end{aligned} \]

(Here \((3 + 2i)^{3} = -9 + 46i\).) So \(z_{1}\) is real and \(z_{2}\) purely imaginary, which also makes \(z_{1} + z_{2} = 88 + 92i\) not real, ruling out the last option.

Answer: A

The value is -16.

Let \(w = z_{1}\overline{z_{2}} + z_{2}\overline{z_{1}}\). Its two terms are conjugates of each other, so \(w = 2\,\text{Re}(z_{1}\overline{z_{2}})\) is real, and \(\overline{w} = w\).

\[ \begin{aligned} z_{1}\overline{z_{2}} &= (1 + 3i)(5 + 3i) \\ &= 5 + 3i + 15i - 9 = -4 + 18i \\ w &= 2(-4) = -8 \\ w + \overline{w} &= -16 \end{aligned} \]

Because \(w\) is real, the answer cannot contain \(i\), which rules out 1+\(i\), 1-\(i\) and -16\(i\).

Answer: E

The value is 2.

The modulus of a quotient is the quotient of the moduli. Each of \(\cos\alpha \pm i\sin\alpha\) and \(\sin\alpha \pm i\cos\alpha\) has modulus

\[ \sqrt{\cos^{2}\alpha + \sin^{2}\alpha} = 1 \]

So each fraction has modulus \(\frac{1}{1} = 1\), and any power of 1 is 1. The sum is \(1^{1000} + 1^{2000} = 1 + 1 = 2\), whatever \(\alpha\) is. The exponents 1000 and 2000 are there to tempt you into expanding with De Moivre's theorem, which is unnecessary.

Answer: E

\(x\) lies in \((-\infty, -2) \cup [0, 5]\).

Bring everything to one side and factor out \(x - 5\):

\[ \begin{aligned} \frac{2(x - 5)}{x + 2} - (x - 5) &\geq 0 \\ \frac{(x - 5)(2 - x - 2)}{x + 2} &\geq 0 \\ \frac{x(x - 5)}{x + 2} &\leq 0 \end{aligned} \]

The critical points are -2, 0 and 5. Testing the intervals: the expression is negative for \(x \lt -2\), positive between -2 and 0, zero or negative on \([0, 5]\), and positive beyond 5. \(x = -2\) is excluded because the denominator vanishes. So the solution is \((-\infty, -2) \cup [0, 5]\).

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.