Complex Numbers & Quadratic Equations
1. Value of \(\left(\frac{2i}{1+i}\right)^2\) is
Solution \(\frac{2i}{1+i} \times \frac{1-i}{1-i} = \frac{2i - 2i^2}{1 - i^2} = \frac{2i+2}{2} = 1+i\).
Now, \((1+i)^2 = 1 + i^2 + 2i = 1 - 1 + 2i = 2i\).
Now, \((1+i)^2 = 1 + i^2 + 2i = 1 - 1 + 2i = 2i\).
2. If \(\left(\frac{1-i}{1+i}\right)^{100} = a + ib\), then
Solution \(\frac{1-i}{1+i} = \frac{(1-i)^2}{1-i^2} = \frac{1-1-2i}{2} = -i\).
\((-i)^{100} = i^{100} = (i^4)^{25} = 1\).
So \(a+ib = 1 + 0i \Rightarrow a=1, b=0\).
\((-i)^{100} = i^{100} = (i^4)^{25} = 1\).
So \(a+ib = 1 + 0i \Rightarrow a=1, b=0\).
3. \(1 + i^2 + i^4 + i^6 + \dots + i^{2n}\) is
Solution The series is \(1 - 1 + 1 - 1 \dots\). The sum depends on the number of terms (\(n+1\)). If even terms, sum is 0. If odd terms, sum is 1. Thus, cannot be uniquely determined without n.
4. If \((x+iy)(2-3i) = 4+i\), then
Solution \(x+iy = \frac{4+i}{2-3i} \times \frac{2+3i}{2+3i} = \frac{8+12i+2i-3}{4+9} = \frac{5+14i}{13}\).
\(x = 5/13, y=14/13\).
\(x = 5/13, y=14/13\).
5. If \(4x + i(3x-y) = 3 + i(-6)\), where x and y are real numbers, then x and y are
Solution Compare real parts: \(4x = 3 \Rightarrow x = 3/4\).
Compare imaginary parts: \(3x - y = -6 \Rightarrow 3(3/4) - y = -6 \Rightarrow y = 9/4 + 6 = 33/4\).
Compare imaginary parts: \(3x - y = -6 \Rightarrow 3(3/4) - y = -6 \Rightarrow y = 9/4 + 6 = 33/4\).
6. If \(z = x-iy\) and \(z^{1/3} = p+iq\), then \(\left(\frac{x}{p} + \frac{y}{q}\right)/(p^2+q^2)\) is equal to
Solution \(x-iy = (p+iq)^3 = p^3 - 3pq^2 + i(3p^2q - q^3)\).
\(x = p(p^2-3q^2), -y = q(3p^2-q^2) \Rightarrow y = q(q^2-3p^2)\).
\(\frac{x}{p} + \frac{y}{q} = p^2-3q^2 + q^2-3p^2 = -2(p^2+q^2)\). Ratio is -2.
\(x = p(p^2-3q^2), -y = q(3p^2-q^2) \Rightarrow y = q(q^2-3p^2)\).
\(\frac{x}{p} + \frac{y}{q} = p^2-3q^2 + q^2-3p^2 = -2(p^2+q^2)\). Ratio is -2.
7. The polar form of the complex number \((i^{25})^3\) is
Solution \(i^{25} = i\). Then \((i)^3 = -i\).
Polar form of \(-i\) is \(r=1, \theta=-\pi/2\) or \(3\pi/2\).
Option (b) is \(\cos(\pi/2) - i\sin(\pi/2) = 0 - i = -i\).
Polar form of \(-i\) is \(r=1, \theta=-\pi/2\) or \(3\pi/2\).
Option (b) is \(\cos(\pi/2) - i\sin(\pi/2) = 0 - i = -i\).
8. If \(z_1 = \sqrt{3} + i\sqrt{3}\) and \(z_2 = \sqrt{3} + i\), then in which quadrant \(\left(\frac{z_1}{z_2}\right)\) lies?
Solution Argument \(\theta_1 = \tan^{-1}(1) = \pi/4\). Argument \(\theta_2 = \tan^{-1}(1/\sqrt{3}) = \pi/6\).
Arg\((z_1/z_2) = \pi/4 - \pi/6 = \pi/12\). Since angle is positive and acute, it lies in Quadrant I.
Arg\((z_1/z_2) = \pi/4 - \pi/6 = \pi/12\). Since angle is positive and acute, it lies in Quadrant I.
9. The solutions of quadratic equation \(ax^2+bx+c=0\), where \(b^2-4ac < 0\), are given by
Solution The standard formula for complex roots when \(D<0\) is \(x = \frac{-b \pm i\sqrt{|D|}}{2a}\), where \(|D| = 4ac - b^2\). Option (a) matches this form.
10. If \(x^2 + x + 1 = 0\), then what is the value of x?
Solution Using quadratic formula: \(x = \frac{-1 \pm \sqrt{1-4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm \sqrt{3}i}{2}\). These are the cube roots of unity (\(\omega, \omega^2\)).
11. The solution of \(\sqrt{3x^2 - 2} = 2x - 1\) are:
Solution Squaring both sides: \(3x^2 - 2 = (2x-1)^2 = 4x^2 - 4x + 1\).
\(x^2 - 4x + 3 = 0 \Rightarrow (x-3)(x-1) = 0\).
Roots are \(x=1, 3\). Checking validity: both satisfy the original equation.
\(x^2 - 4x + 3 = 0 \Rightarrow (x-3)(x-1) = 0\).
Roots are \(x=1, 3\). Checking validity: both satisfy the original equation.
12. If \(\alpha, \beta\) are roots of \(x^2 - 5x + 6 = 0\), then equation with roots \(\alpha+3\) and \(\beta+3\) is:
Solution Roots of \(x^2 - 5x + 6 = 0\) are 2 and 3.
New roots: \(2+3=5\) and \(3+3=6\).
Equation: \(x^2 - (5+6)x + (5)(6) = 0 \Rightarrow x^2 - 11x + 30 = 0\).
New roots: \(2+3=5\) and \(3+3=6\).
Equation: \(x^2 - (5+6)x + (5)(6) = 0 \Rightarrow x^2 - 11x + 30 = 0\).
13. Value of k such that equations \(2x^2 + kx - 5 = 0\) and \(x^2 - 3x - 4 = 0\) have one common root is
Solution Roots of \(x^2 - 3x - 4 = 0\) are 4, -1.
If 4 is common: \(2(16) + 4k - 5 = 0 \Rightarrow 4k = -27 \Rightarrow k = -27/4\).
If -1 is common: \(2(1) - k - 5 = 0 \Rightarrow k = -3\).
If 4 is common: \(2(16) + 4k - 5 = 0 \Rightarrow 4k = -27 \Rightarrow k = -27/4\).
If -1 is common: \(2(1) - k - 5 = 0 \Rightarrow k = -3\).
14. If \(a < b < c < d\), then nature of roots of \((x-a)(x-c) + 2(x-b)(x-d) = 0\) is
Solution Let \(f(x) = (x-a)(x-c) + 2(x-b)(x-d)\).
\(f(b) = (b-a)(b-c) + 0 < 0\).
\(f(d) = (d-a)(d-c) + 0 > 0\).
Since sign changes between roots, roots are real and distinct.
\(f(b) = (b-a)(b-c) + 0 < 0\).
\(f(d) = (d-a)(d-c) + 0 > 0\).
Since sign changes between roots, roots are real and distinct.
15. For equation \(\frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{x+c}\), if product of roots is zero, sum of roots is
Solution Simplification leads to \(x^2 + 2cx + (ac+bc-ab) = 0\).
Prod=0 \(\Rightarrow ac+bc=ab \Rightarrow c = ab/(a+b)\).
Sum of roots = \(-2c = -2ab/(a+b)\). Note: Image options are permuted variables, but logic holds.
Prod=0 \(\Rightarrow ac+bc=ab \Rightarrow c = ab/(a+b)\).
Sum of roots = \(-2c = -2ab/(a+b)\). Note: Image options are permuted variables, but logic holds.
16. Product of real roots of the equation \(t^2x^2 + |x| + 9 = 0\)
Solution \(t^2x^2 \ge 0\), \(|x| \ge 0\), \(9 > 0\). Sum of positive terms cannot be zero.
Thus, no real roots exist. Product does not exist.
Thus, no real roots exist. Product does not exist.
17. If p and q are roots of \(x^2 + px + q = 0\), then
Solution Sum \(p+q = -p \Rightarrow 2p = -q\). Product \(pq = q\).
If \(q \neq 0, p=1\). Then \(2(1) = -q \Rightarrow q=-2\).
Solution: \(p=1, q=-2\).
If \(q \neq 0, p=1\). Then \(2(1) = -q \Rightarrow q=-2\).
Solution: \(p=1, q=-2\).
18. The roots of equation \((p-q)x^2 + (q-r)x + (r-p) = 0\) are
Solution Sum of coefficients is 0, so 1 is a root.
Product of roots is \(\frac{r-p}{p-q}\). So the roots are 1 and \(\frac{r-p}{p-q}\).
Product of roots is \(\frac{r-p}{p-q}\). So the roots are 1 and \(\frac{r-p}{p-q}\).
19. If \(\alpha, \beta\) are roots of \(ax^2+bx+c=0\), then \(\alpha \beta^2 + \alpha^2 \beta + \alpha \beta\) equals
Solution Expression = \(\alpha\beta(\beta + \alpha + 1) = \frac{c}{a}(\frac{-b}{a} + 1) = \frac{c}{a}(\frac{a-b}{a}) = \frac{c(a-b)}{a^2}\).
20. The roots of equation \(x - \frac{2}{x-1} = 1 - \frac{2}{x-1}\) is
Solution Canceling terms implies \(x=1\). However, at \(x=1\), the term \(\frac{2}{x-1}\) is undefined.
Therefore, the solution set is empty (zero roots). Correct choice is None of these (meaning 0 solutions).
Therefore, the solution set is empty (zero roots). Correct choice is None of these (meaning 0 solutions).