Complex Numbers and Quadratic Equations Questions and Answers

Answer: C

x lies in \((-\infty, 1) \cup (3, \infty)\).

Do not multiply through by \(x - 1\), whose sign is unknown. Bring everything to one side instead:

\[ \begin{aligned} \frac{x + 1}{x - 1} - 2 &\lt 0 \\ \frac{3 - x}{x - 1} &\lt 0 \\ \frac{x - 3}{x - 1} &\gt 0 \end{aligned} \]

The quotient is positive when numerator and denominator have the same sign: x greater than 3, or x less than 1. (x = 1 is excluded, where the expression is undefined.) The interval (1, 3) is where the inequality fails, and the options with -1 or -3 come from sign slips in the rearrangement.

Answer: B

The solution set is \(-5 \lt x \lt 1\).

\(|u| \lt 3\) means \(-3 \lt u \lt 3\). With \(u = x + 2\):

\[ \begin{aligned} -3 &\lt x + 2 \lt 3 \\ -5 &\lt x \lt 1 \end{aligned} \]

Geometrically, x is within 3 units of -2. The option \(-1 \lt x \lt 5\) is the solution of \(|x - 2| \lt 3\), the usual sign slip. The other intervals do not have -2 at their centre.

Answer: E

The value is 10.

Square \(z = 2 + i\) first, using \(i^{2} = -1\):

\[ \begin{aligned} z^{2} &= 4 + 4i + i^{2} = 3 + 4i \\ z^{2} - 4z + 15 &= (3 + 4i) - (8 + 4i) + 15 \\ &= 10 \end{aligned} \]

The imaginary parts cancel exactly. A quicker route: \(z - 2 = i\), so \((z-2)^{2} = -1\), that is \(z^{2} - 4z + 4 = -1\), and adding 11 gives 10.

15 is what you get by forgetting that \(z^{2} - 4z\) is not zero; 6 and 12 come from treating \(i^{2}\) as \(+1\) at some step.

Answer: E

The modulus is \(\frac{5}{2}\).

Simplify \(\frac{2}{i}\) first: multiplying top and bottom by \(i\) gives \(\frac{2i}{i^{2}} = -2i\). Then:

\[ \begin{aligned} \frac{i}{2} - \frac{2}{i} &= \frac{i}{2} + 2i \\ &= \frac{5i}{2} \\ \left|\frac{5i}{2}\right| &= \frac{5}{2} \end{aligned} \]

The number is purely imaginary, so its modulus is just the size of the coefficient of \(i\). Writing \(\frac{2}{i}\) as \(+2i\) instead of \(-2i\) gives \(\frac{i}{2} - 2i = -\frac{3i}{2}\) and the distractor \(\frac{3}{2}\).

Answer: B

The locus is a straight line.

Write \(z = x + iy\). Then \(z - 2 - 3i = (x - 2) + i(y - 3)\), and equal real and imaginary parts means:

\[ \begin{aligned} x - 2 &= y - 3 \\ x - y + 1 &= 0 \end{aligned} \]

This is a first-degree equation in \(x\) and \(y\), so the point \(z\) moves on a line of slope 1 through \((2, 3)\).

A circle, parabola, ellipse or hyperbola needs a second-degree equation; that would arise if the condition involved the modulus of \(z - 2 - 3i\) or a product of its parts, not a plain equality of the two parts.

Answer: E

\(i^{k} = -i\).

Powers of \(i\) repeat every four: \(i^{1} = i,\ i^{2} = -1,\ i^{3} = -i,\ i^{4} = 1\). Since \(i^{4n} = (i^{4})^{n} = 1\) for any integer \(n\):

\[ \begin{aligned} i^{4n+3} &= i^{4n} \cdot i^{3} \\ &= 1 \times (-i) = -i \end{aligned} \]

The other values belong to the other remainders: exponent \(4n\) gives 1, \(4n + 1\) gives \(i\), \(4n + 2\) gives \(-1\). Zero is never a power of \(i\).

Answer: C

\(x\) lies in \((5, 6)\).

Do not multiply across by \(x - 5\), whose sign is unknown. Bring everything to one side instead:

\[ \begin{aligned} \frac{x - 3}{x - 5} - 3 &\gt 0 \\ \frac{x - 3 - 3x + 15}{x - 5} &\gt 0 \\ \frac{2(6 - x)}{x - 5} &\gt 0 \end{aligned} \]

The fraction is positive when the numerator and denominator have the same sign: both positive needs \(x \lt 6\) and \(x \gt 5\); both negative needs \(x \gt 6\) and \(x \lt 5\), impossible. So \(5 \lt x \lt 6\).

Check \(x = 5.5\): \(2.5/0.5 = 5 \gt 3\). Options (3,8) and (5,8) include points such as 7, where the fraction is \(4/2 = 2\), not greater than 3.

Answer: A

The solution set is \(\left(-\infty, -\frac{1}{2}\right) \cup \left(\frac{1}{6}, \infty\right)\).

Remove the modulus, then solve for \(\frac{1}{x}\):

\[ \begin{aligned} -4 \lt \frac{1}{x} - 2 &\lt 4 \\ -2 \lt \frac{1}{x} &\lt 6 \end{aligned} \]

For \(x \gt 0\), \(\frac{1}{x}\) is positive, so only \(\frac{1}{x} \lt 6\) matters, giving \(x \gt \frac{1}{6}\). For \(x \lt 0\), \(\frac{1}{x}\) is negative, so only \(\frac{1}{x} \gt -2\) matters, giving \(x \lt -\frac{1}{2}\).

The single intervals in the next two options each keep only one of these two branches. The fourth option uses \(\frac{1}{2}\) where \(\frac{1}{6}\) belongs and wrongly includes small positive \(x\) such as 0.1, where \(|10 - 2| = 8\).

Answer: D

\(x + y = -1\).

Simplify the right side by multiplying by the conjugate of \(1 + i\):

\[ \begin{aligned} \frac{7 - 7i}{1 + i} &= \frac{(7 - 7i)(1 - i)}{2} \\ &= \frac{7 - 14i + 7i^{2}}{2} = -7i \end{aligned} \]

With \(z = x + iy\) and \(\bar{z} = x - iy\):

\[ \begin{aligned} 5z - 2\bar{z} &= 3x + 7iy \\ 3x + 7iy &= -7i \end{aligned} \]

Equating real and imaginary parts gives \(x = 0\) and \(y = -1\), so \(x + y = -1\). Check: with \(z = -i\), \(5(-i) - 2(i) = -7i\), as required.

Answer: B

\(\lambda = 13\).

With \(z = 1 + i\): \(\bar{z} = 1 - i\) and \(z^{2} = 1 + 2i + i^{2} = 2i\).

\[ \begin{aligned} \frac{24\bar{z}}{z^{2}} &= \frac{24(1 - i)}{2i} = \frac{12(1 - i)(-i)}{1} \\ &= 12(-i + i^{2}) = -12 - 12i \end{aligned} \]

\[ \begin{aligned} z - \frac{24\bar{z}}{z^{2}} &= 1 + i + 12 + 12i \\ &= 13 + 13i = 13(1 + i) \end{aligned} \]

So the expression equals \(13z\) and \(\lambda = 13\). The option 12 forgets the original \(z\) that is added back.