Complex Numbers and Quadratic Equations Questions and Answers

Answer: C

The correct match is (P) → (1), (Q) → (2), (R) → (4), (S) → (5).

The roots of \(x^{2} + x + 1 = 0\) are the cube roots of unity \(\omega, \omega^{2}\), with \(\omega^{3} = 1\) and \(1 + \omega + \omega^{2} = 0\). So \(\omega + 1 = -\omega^{2}\) and \(\omega^{2} + 1 = -\omega\).

(P) \((\omega + 1)^{2026} = \omega^{4052} = \omega^{2}\) (since \(4052 = 3 \times 1350 + 2\)), and \((\omega^{2} + 1)^{2026} = \omega^{2026} = \omega\). The roots are the reciprocals \(\omega\) and \(\omega^{2}\), so the equation is \(x^{2} + x + 1 = 0\).

(Q) With the odd power 2027 a minus sign survives: \((-\omega^{2})^{2027} = -\omega\) and \((-\omega)^{2027} = -\omega^{2}\). The roots are \(-\omega^{2}\) and \(-\omega\), with sum 1 and product 1, so the equation is \(x^{2} - x + 1 = 0\).

(R) The roots of \(x^{2} - x + 1 = 0\) are \(-\omega\) and \(-\omega^{2}\), so \(\gamma - 1 = \omega^{2}\) and \(\delta - 1 = \omega\). The sum is \(\frac{1}{\omega^{4052}} + \frac{1}{\omega^{2026}} = \omega + \omega^{2} = -1\).

(S) Put \(a = p + 1\), \(b = r + 1\). They satisfy \(y^{2} - y - 1 = 0\), so \(a + b = 1\) and \(ab = -1\):

\[ \begin{aligned} \frac{1}{a^{3}} + \frac{1}{b^{3}} &= \frac{(a + b)^{3} - 3ab(a + b)}{(ab)^{3}} \\ &= \frac{1 + 3}{-1} = -4 \end{aligned} \]

Only the third option has all four pairings. The first swaps R and S, the fourth also gets P and Q wrong, and the second gets P and Q wrong.

Answer: B

\(z^{4} = \) -4.

Since \(\frac{1}{i} = \frac{i}{i^{2}} = -i\), we have \(z = 1 - i\). Then

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

Taking \(\frac{1}{i} = i\) by mistake gives \(z = 1 + i\), whose fourth power is also -4, but the options \(1 - i\) and \(1 + i\) are just \(z\) itself or its conjugate, not \(z^{4}\). The value 4 comes from dropping the sign of \(i^{2}\).

Answer: E

The modulus is 16.

The modulus of a power is the power of the modulus, \(|w^{2}| = |w|^{2}\). Here \(w = 2\sqrt{2} + 2\sqrt{2}\,i\):

\[ \begin{aligned} |w| &= \sqrt{(2\sqrt{2})^{2} + (2\sqrt{2})^{2}} = \sqrt{8 + 8} = 4 \\ |w^{2}| &= 4^{2} = 16 \end{aligned} \]

4 is \(|w|\) without squaring; 64 is \(|w|^{3}\); 8 and 32 come from adding the parts instead of squaring them.

Answer: D

\(|z|^{2} = \) 13.

With \(z = x + iy\), \(z + \bar{z} = 2x\) and \(z - \bar{z} = 2iy\). So \(2x = 6\) gives \(x = 3\), and \(2iy = 4i\) gives \(y = 2\).

\[ |z|^{2} = x^{2} + y^{2} = 9 + 4 = 13 \]

36 and 16 square the sums 6 and 4 instead of the parts; 9 is \(x^{2}\) alone; 15 does not arise from any of the parts.

Answer: D

\(\alpha = \) 2.

Multiply numerator and denominator by the conjugate \(\alpha - i\):

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

So \(Re(z) = \frac{2\alpha - 1}{\alpha^{2} + 1}\) and \(Im(\bar{z}) = -Im(z) = \frac{2 + \alpha}{\alpha^{2} + 1}\). The condition gives

\[ \begin{aligned} 4(2\alpha - 1) &= 3(2 + \alpha) \\ 8\alpha - 4 &= 6 + 3\alpha \\ \alpha &= 2 \end{aligned} \]

Using \(Im(z)\) instead of \(Im(\bar{z})\) flips the sign of the right side and leads to a non-integer value, so none of -2, 3, 5 or -5 satisfies the stated condition.

Answer: D

The solution set is \(\left(\frac{-5}{3}, 2\right]\).

Subtract 2 throughout, then divide by -3, which reverses both inequality signs:

\[ \begin{aligned} -4 \leq 2 - 3x &\lt 7 \\ -6 \leq -3x &\lt 5 \\ 2 \geq x &\gt -\tfrac{5}{3} \end{aligned} \]

So \(x\) runs from \(-\frac{5}{3}\) (excluded) to 2 (included). Intervals such as \(\left(2, \frac{5}{3}\right)\) come from forgetting to reverse the signs, which produces an empty, back-to-front interval; the -11/3 and -7/3 endpoints come from arithmetic slips.

Answer: E

The least value of \(b\) is -1.

Multiply the given range by 5 and add 4:

\[ \begin{aligned} -5 \lt x &\leq -1 \\ -25 \lt 5x &\leq -5 \\ -21 \lt 5x + 4 &\leq -1 \end{aligned} \]

\(5x + 4\) reaches -1 (at \(x = -1\)), so any valid upper bound \(b\) must be at least -1, and the least is -1 itself. -4 and -5 are too small (they fail at \(x = -1\)); 4 and 5 are valid bounds but not the least.

Answer: D

\((1 + i)^{10} =\) \(32i\).

Square first, since \((1 + i)^{2}\) is simple:

\[ \begin{aligned} (1 + i)^{2} &= 1 + 2i + i^{2} = 2i \\ (1 + i)^{10} &= (2i)^{5} = 32\,i^{5} = 32i \end{aligned} \]

because \(i^{5} = i^{4}\cdot i = i\). The modulus check agrees: \(|1 + i|^{10} = (\sqrt{2})^{10} = 32\). The options 16 and \(16i\) have the wrong modulus, and 32 or 64 lose the factor \(i\).

Answer: B

The sum is 1.

The powers of \(i\) repeat every four terms, and any four consecutive powers add to zero: \(i^{k} + i^{k+1} + i^{k+2} + i^{k+3} = i^{k}(1 + i - 1 - i) = 0\).

From \(i^{3}\) to \(i^{93}\) there are 91 terms. The first 88 (\(i^{3}\) to \(i^{90}\)) form 22 complete groups and cancel, leaving

\[ i^{91} + i^{92} + i^{93} = -i + 1 + i = 1 \]

The other options come from miscounting the leftover terms: leaving one or two terms gives \(-i\) or \(1 - i\), and a whole number of cycles gives 0.

Answer: C

The imaginary part is 3.

Simplify the fraction first by multiplying top and bottom by \(-i\):

\[ \begin{aligned} \frac{3 - 2i}{2i} &= \frac{(3 - 2i)(-i)}{2} = \frac{-2 - 3i}{2} \\ \left(\frac{-2 - 3i}{2}\right)^{2} &= \frac{4 + 12i + 9i^{2}}{4} \\ &= \frac{-5 + 12i}{4} = -\frac{5}{4} + 3i \end{aligned} \]

So the imaginary part is 3. \(-\frac{5}{4}\) is the real part, which is the trap; \(\frac{5}{4}\), \(-3\) and \(\frac{3}{4}\) come from sign or scaling slips.