Samacheer Kalvi Class 12 Maths Solutions Chapter 2 Complex Numbers Exercise 2.9

Choose The Most Suitable Answer.

Question 1. \( i^n + i^{n+1} + i^{n+2} + i^{n+3} \) is:
(a) 0
(b) 1
(c) -1
(d) z
Answer: (a) 0
In simple words: The sum of any four consecutive powers of \( i \) (like \( i^n, i^{n+1}, i^{n+2}, i^{n+3} \)) is always zero. This is because \( i^n \) can be factored out, leaving \( 1 + i + i^2 + i^3 \), which simplifies to \( 1 + i - 1 - i = 0 \).

🎯 Exam Tip: Remember the cycle of powers of \( i \): \( i^1=i, i^2=-1, i^3=-i, i^4=1 \). This cycle repeats every four powers, making the sum of any four consecutive powers zero.

Question 2. The value of \( \sum _{ i=1 }^{ 13 }{ ({ i }^{ n }+{ i }^{ n-1 }) } \) is
(a) 1 + i
(b) i
(c) 1
(d) 0
Answer: (a) 1 + i
In simple words: This sum is a series of complex numbers. The hint shows that when you sum these terms, they group into repeating sets of powers of \( i \), which mostly cancel out to zero. After the cancellations, you are left with just \( 1 + i \). The cyclic nature of powers of \( i \) simplifies these types of sums greatly.

🎯 Exam Tip: When dealing with sums of powers of \( i \), always look for groups of four terms as they sum to zero. This helps simplify complex series quickly.

Question 3. The area of the triangle formed by the complex numbers z, iz, and z + iz in the Argand's diagram is:
(a) \( \frac{1}{2} |z|^2 \)
(b) \( |z|^2 \)
(c) \( \frac{3}{2} |z|^2 \)
Answer: (a) \( \frac{1}{2} |z|^2 \)
In simple words: Imagine these complex numbers as points on a graph. The numbers \( z \), \( iz \), and \( z+iz \) form a right-angled triangle. The area of such a triangle is half the product of its base and height. In this case, the lengths of the sides originating from the origin are \( |z| \) and \( |iz| \) which is also \( |z| \). This makes it an isosceles right-angled triangle, and its area is \( \frac{1}{2} |z|^2 \).

🎯 Exam Tip: When complex numbers \( z \) and \( iz \) represent vertices of a triangle, it often indicates a right angle at the origin because multiplication by \( i \) rotates a complex number by 90 degrees.

Question 4. If the conjugate of a complex number is \( \frac{1}{i-2} \), then the complex number is:
(a) \( \frac{1}{i+2} \)
(b) \( \frac{-1}{i+2} \)
(c) \( \frac{-1}{-2} \)
(d) \( \frac{1}{-2} \)
Answer: (b) \( \frac{-1}{i+2} \)
In simple words: To find the original complex number when you have its conjugate, you just take the conjugate of the conjugate. This means you change the sign of every 'i' term. When you take the conjugate of \( \frac{1}{i-2} \), it becomes \( \frac{1}{-i-2} \), which can be written as \( \frac{-1}{i+2} \).

🎯 Exam Tip: The conjugate of a fraction \( \frac{a+bi}{c+di} \) is \( \frac{a-bi}{c-di} \). Remember to apply the conjugate operation to both the numerator and the denominator.

Question 5. If \( z = \frac{(\sqrt{3}+i)^3(3i+4)^2}{(8+6i)^2} \), then the value of \( |z| \) is:
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (c) 2
In simple words: To find the absolute value (or modulus) of a complex number like this, you can use the property that the modulus of a product is the product of the moduli, and the modulus of a quotient is the quotient of the moduli. This means you find the modulus of each part of the fraction separately and then do the multiplication and division. The calculations for \( |\sqrt{3}+i| \), \( |3i+4| \), and \( |8+6i| \) simplify nicely.

🎯 Exam Tip: Remember the modulus property \( |\frac{z_1 z_2}{z_3}| = \frac{|z_1| |z_2|}{|z_3|} \). This simplifies calculations immensely when dealing with complex fractions and powers.

Question 6. If z is a non zero complex number, such that \( 2iz^2 = \bar { z } \) then \( |z| \) is:
(a) \( \frac{1}{2} \)
(b) 0
(c) 1
(d) 2
Answer: (a) \( \frac{1}{2} \)
In simple words: This problem involves finding the magnitude of a complex number z from an equation relating \( z \) and its conjugate \( \bar{z} \). A key step is to take the modulus of both sides of the equation. Since \( |2i| = 2 \) and \( |z^2| = |z|^2 \), and \( |\bar{z}| = |z| \), the equation simplifies to \( 2|z|^2 = |z| \). Since \( z \) is non-zero, \( |z| \) is not zero, so you can divide by \( |z| \) to find its value.

🎯 Exam Tip: When an equation involves both a complex number \( z \) and its conjugate \( \bar{z} \), taking the modulus of both sides is often the first step to find \( |z| \).

Question 7. If \( |z - 2 + i | \le 2 \), then the greatest value of \( |z| \) is:
(a) \( \sqrt{3}-2 \)
(b) \( \sqrt{3} + 2 \)
(c) \( \sqrt{5}-2 \)
(d) \( \sqrt{5} + 2 \)
Answer: (d) \( \sqrt{5} + 2 \)
In simple words: This question asks for the maximum possible distance of a complex number \( z \) from the origin, given a condition. The condition \( |z - (2 - i)| \le 2 \) describes all points \( z \) inside or on a circle centered at \( (2, -1) \) with a radius of \( 2 \). To find the maximum distance \( |z| \) from the origin, you add the distance from the origin to the center of the circle \( |2 - i| \) and the radius. The distance from the origin to \( (2,-1) \) is \( \sqrt{2^2 + (-1)^2} = \sqrt{5} \). Adding the radius gives \( \sqrt{5} + 2 \).

🎯 Exam Tip: The minimum and maximum values of \( |z| \) for a complex number \( z \) satisfying \( |z-c| = r \) (a circle) are \( ||c|-r| \) and \( |c|+r \) respectively. For \( |z-c| \le r \) (a disk), the minimum is 0 if origin is inside, or \( |c|-r \) if origin is outside, and maximum is \( |c|+r \).

Question 8. If \( |z - 1| = 2 \), then the least value of \( |z| \) is:
(a) 1
(b) 2
(c) 3
(d) 5
Answer: (a) 1
In simple words: The equation \( |z - 1| = 2 \) describes a circle on the complex plane. This circle is centered at the point \( (1, 0) \) and has a radius of \( 2 \). We want to find the smallest possible distance \( |z| \) from the origin to any point on this circle. The points closest to the origin are found by subtracting the radius from the distance of the center to the origin. The center is at \( (1,0) \), so its distance from the origin is \( |1| = 1 \). Subtracting the radius \( 2 \) gives \( |1 - 2| = |-1| = 1 \). This is the minimum distance.

🎯 Exam Tip: For a circle \( |z-c| = r \), the least value of \( |z| \) is \( ||c|-r| \) and the greatest value is \( |c|+r \). Applying this formula correctly is key.

Question 9. If \( |z| = 1 \), then the value of \( \frac { 1+z }{ 1+\bar { z } } \) is
(a) z
(b) \( \bar{z} \)
(c) \( \frac{1}{z} \)
(d) 1
Answer: (a) z
In simple words: Since \( |z| = 1 \), we know that \( z \times \bar{z} = 1 \). This means that \( \bar{z} \) can be replaced by \( \frac{1}{z} \). If you substitute \( \frac{1}{z} \) for \( \bar{z} \) in the denominator of the expression, you can then simplify the fraction by finding a common denominator and multiplying. After simplification, the entire expression becomes just \( z \).

🎯 Exam Tip: The condition \( |z|=1 \) is very useful. Always remember that it implies \( z\bar{z}=1 \) and thus \( \bar{z}=\frac{1}{z} \). This substitution often simplifies complex expressions significantly.

Question 10. The solution of the equation \( |z| - z = 1 + 2i \) is:
(a) \( \frac{3}{2} – 2i \)
(b) \( - \frac{3}{2} + 2i \)
(c) \( 2 - \frac{3}{2}i \)
(d) \( 2 + \frac{3}{2}i \)
Answer: (a) \( \frac{3}{2} – 2i \)
In simple words: To solve this equation, first let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then, \( |z| \) becomes \( \sqrt{x^2+y^2} \). Substitute these into the given equation, then separate the real and imaginary parts on both sides. This will give you two equations with \( x \) and \( y \), which you can solve simultaneously to find the values of \( x \) and \( y \), and thus \( z \). It's important to carefully equate the real and imaginary components.

🎯 Exam Tip: When solving equations involving complex numbers and their moduli, always substitute \( z = x+iy \) and then equate the real parts and imaginary parts on both sides of the equation. This transforms the complex equation into a system of real equations.

Question 11. If \( |z_1| = 1, |z_2| = 2, |z_3| = 3 \) and \( |9z_1z_2 + 4z_1z_3 + z_2z_3| = 12 \), then the value of \( |z_1 + z_2 + z_3| \) is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
In simple words: This problem uses the property that for any complex number \( z \), \( |z|^2 = z\bar{z} \). You can rewrite the term \( |9z_1z_2 + 4z_1z_3 + z_2z_3| \) by taking out common factors or manipulating it using the given moduli. The key is to notice that \( \frac{1}{\bar{z}_k} = z_k \) if \( |z_k|=1 \) (and similar for other moduli) and then simplify the expression to relate it to \( |z_1 + z_2 + z_3| \).

🎯 Exam Tip: When dealing with moduli of sums or products, remember the properties \( |z_1z_2| = |z_1||z_2| \) and \( |\bar{z}| = |z| \). The condition \( |z|=k \implies z\bar{z}=k^2 \) is crucial for manipulating such expressions.

Question 12. If z is a complex number such that \( z \in C \setminus R \) (z is complex but not real) and \( z + \frac{1}{z} \in R \) ( \( z + \frac{1}{z} \) is real), then \( |z| \) is:
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (b) 1
In simple words: If \( z + \frac{1}{z} \) is a real number, it means its imaginary part is zero. Let \( z = x + iy \). When you calculate \( z + \frac{1}{z} \), the imaginary terms must cancel out. After doing the algebra, this condition forces \( y(1 - (x^2 + y^2)) = 0 \). Since \( z \) is not real, \( y \ne 0 \), so we must have \( 1 - (x^2 + y^2) = 0 \). Because \( |z|^2 = x^2 + y^2 \), this means \( 1 - |z|^2 = 0 \), which simplifies to \( |z|^2 = 1 \), so \( |z|=1 \).

🎯 Exam Tip: A complex number is real if and only if it is equal to its conjugate, i.e., \( Z = \bar{Z} \). Use this property to set up an equation for \( z + \frac{1}{z} \) and solve for \( |z| \).

Question 13. \( z_1, z_2 \) and \( z_3 \) are complex numbers such that \( z_1 + z_2 + z_3 = 0 \) and \( |z_1| = |z_2| = |z_3| = 1 \) then \( z_1^2 + z_2^2 + z_3^2 \) is:
(a) 3
(b) 2
(c) 1
(d) 0
Answer: (d) 0
In simple words: Since \( |z_k| = 1 \) for each \( z_k \), we know that \( z_k \bar{z}_k = 1 \), which means \( \bar{z}_k = \frac{1}{z_k} \). If \( z_1 + z_2 + z_3 = 0 \), then taking the conjugate of this equation gives \( \bar{z}_1 + \bar{z}_2 + \bar{z}_3 = 0 \). Substituting \( \frac{1}{z_k} \) for \( \bar{z}_k \), we get \( \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} = 0 \). Multiplying by \( z_1z_2z_3 \) gives \( z_2z_3 + z_1z_3 + z_1z_2 = 0 \). Now consider \( (z_1+z_2+z_3)^2 = z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1) \). Since \( z_1+z_2+z_3=0 \) and \( z_1z_2 + z_2z_3 + z_3z_1 = 0 \), it follows that \( z_1^2 + z_2^2 + z_3^2 \) must also be zero.

🎯 Exam Tip: This is a standard result for complex numbers on the unit circle that sum to zero. The identities \( |z|=1 \implies \bar{z}=\frac{1}{z} \) and \( (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \) are crucial here.

Question 14. If \( \frac{z-1}{z+1} \) is purely imaginary, then \( |z| \) is
(a) \( \frac{1}{2} \)
(b) 1
(c) 2
(d) 3
Answer: (b) 1
In simple words: If a complex number is purely imaginary, its real part must be zero. Let \( z = x + iy \). Substitute this into the expression \( \frac{z-1}{z+1} \). After simplifying the fraction to the form \( A+Bi \), set the real part \( A \) equal to zero. This will lead to the equation \( x^2 + y^2 - 1 = 0 \). Since \( |z|^2 = x^2 + y^2 \), this means \( |z|^2 = 1 \), so \( |z|=1 \). Geometrically, this means \( z \) lies on the unit circle.

🎯 Exam Tip: A complex number \( w \) is purely imaginary if \( w = -\bar{w} \). Use this property: \( \frac{z-1}{z+1} = -\overline{\left(\frac{z-1}{z+1}\right)} \). This simplifies calculations and helps avoid errors with algebraic expansion.

Question 15. If \( z = x + iy \) is a complex number such that \( |z + 2| = |z – 2| \), then the locus of z is:
(a) real axis
(b) imaginary axis
(c) ellipse
(d) circle
Answer: (b) imaginary axis
In simple words: The equation \( |z + 2| = |z – 2| \) means that the distance from \( z \) to \( -2 \) is the same as the distance from \( z \) to \( 2 \). Geometrically, this implies that \( z \) lies on the perpendicular bisector of the line segment connecting \( -2 \) and \( 2 \). This segment lies on the real axis from \( -2 \) to \( 2 \). Its perpendicular bisector is the imaginary axis (where \( x=0 \)). Algebraically, substituting \( z = x + iy \) leads to \( (x+2)^2 + y^2 = (x-2)^2 + y^2 \), which simplifies to \( 4x = -4x \), meaning \( 8x=0 \), so \( x=0 \). This is the equation of the imaginary axis.

🎯 Exam Tip: The locus of \( z \) satisfying \( |z-a| = |z-b| \) is the perpendicular bisector of the line segment joining complex numbers \( a \) and \( b \).

Question 16. The principal argument of \( \frac{3}{-1+i} \) is:
(a) \( \frac{-5\pi}{6} \)
(b) \( \frac{-2}{3} \)
(c) \( \frac{-3\pi}{4} \)
(d) \( \frac{-\pi}{2} \)
Answer: (c) \( \frac{-3\pi}{4} \)
In simple words: First, simplify the complex fraction by multiplying the numerator and denominator by the conjugate of the denominator. This transforms \( \frac{3}{-1+i} \) into \( \frac{3(-1-i)}{(-1)^2+1^2} = \frac{-3-3i}{2} = -\frac{3}{2} - \frac{3}{2}i \). Now, find the argument of this new complex number. Since both the real and imaginary parts are negative, the number lies in the third quadrant. The reference angle \( \alpha = \tan^{-1}\left(\frac{|-3/2|}{|-3/2|}\right) = \tan^{-1}(1) = \frac{\pi}{4} \). In the third quadrant, the principal argument is \( -\pi + \alpha \) or \( \alpha - \pi \), which is \( \frac{\pi}{4} - \pi = -\frac{3\pi}{4} \).

🎯 Exam Tip: Always rationalize the denominator first to express the complex number in the form \( x+iy \). Then, determine the quadrant to correctly calculate the principal argument within the range \( (-\pi, \pi] \).

Question 17. The principal argument of \( (\sin 40^\circ + i \cos 40^\circ)^5 \) is:
(a) \( – 110^\circ \)
(b) \( -70^\circ \)
(c) \( 70^\circ \)
(d) \( 110^\circ \)
Answer: (a) \( – 110^\circ \)
In simple words: To find the argument of \( (\sin 40^\circ + i \cos 40^\circ)^5 \), first convert \( \sin 40^\circ + i \cos 40^\circ \) into the standard polar form \( \cos \theta + i \sin \theta \). Using trigonometric identities, \( \sin 40^\circ = \cos(90^\circ - 40^\circ) = \cos 50^\circ \) and \( \cos 40^\circ = \sin(90^\circ - 40^\circ) = \sin 50^\circ \). So, the base becomes \( \cos 50^\circ + i \sin 50^\circ \). Now, apply De Moivre's theorem: \( (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) \). This gives \( \cos(5 \times 50^\circ) + i \sin(5 \times 50^\circ) = \cos 250^\circ + i \sin 250^\circ \). The principal argument of \( 250^\circ \) is \( 250^\circ - 360^\circ = -110^\circ \).

🎯 Exam Tip: When given \( \sin \alpha + i \cos \alpha \), convert it to \( \cos(90^\circ-\alpha) + i \sin(90^\circ-\alpha) \) before applying De Moivre's theorem to avoid errors in argument calculation.

Question 18. If \( (1 + i) (1 + 2i) (1 + 3i) \ldots (1 + ni) = x + iy \), then \( 2 \cdot 5 \cdot 10 \ldots (1 + n^2) \) is:
(a) 1
(b) i
(c) \( x^2 + y^2 \)
(d) \( 1 + n^2 \)
Answer: (c) \( x^2 + y^2 \)
In simple words: The given equation involves a product of complex numbers, which equals \( x+iy \). The expression we need to find, \( 2 \cdot 5 \cdot 10 \ldots (1 + n^2) \), can be rewritten as \( (1^2+1^2)(1^2+2^2)(1^2+3^2)\ldots(1^2+n^2) \). This is exactly the product of the squared moduli of each term in the original product. Since the modulus of a product is the product of the moduli, and \( |z|^2 = x^2 + y^2 \) if \( z = x+iy \), the value of the expression is \( |x+iy|^2 \), which is \( x^2 + y^2 \). This is a useful shortcut in complex numbers.

🎯 Exam Tip: Remember that \( |z|^2 = x^2+y^2 \) for \( z = x+iy \). For a product of complex numbers \( z_1z_2\ldots z_n \), the square of its modulus is \( |z_1z_2\ldots z_n|^2 = |z_1|^2|z_2|^2\ldots|z_n|^2 \).

Question 19. If \( \omega \ne 1 \) is a cubic root of unity and \( (1 + \omega)^7 = A + B\omega \), then \( (A, B) \) equals:
(a) (1, 0)
(b) (-1, 1)
(c) (0, 1)
(d) (1, 1)
Answer: (d) (1, 1)
In simple words: This problem uses the properties of cube roots of unity. Since \( \omega \ne 1 \) is a cubic root of unity, we know two main things: \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \). From the first property, \( 1 + \omega = -\omega^2 \). Substitute this into the given equation: \( (-\omega^2)^7 = -\omega^{14} \). Since \( \omega^3 = 1 \), we can simplify \( \omega^{14} = \omega^{12} \cdot \omega^2 = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2 \). So, \( (1 + \omega)^7 = -\omega^2 \). From \( 1 + \omega + \omega^2 = 0 \), we know \( -\omega^2 = 1 + \omega \). Therefore, \( A=1 \) and \( B=1 \).

🎯 Exam Tip: The two fundamental properties of cube roots of unity \( \omega \): \( 1+\omega+\omega^2=0 \) and \( \omega^3=1 \) are essential. Always use them to simplify expressions involving powers of \( \omega \).

Question 20. The principal argument of the complex number \( \frac{(1+i \sqrt{3})^{2}}{4 i(1-i \sqrt{3})} \) is:
(a) \( \frac{2\pi}{3} \)
(b) \( \frac{\pi}{6} \)
(c) \( \frac{5\pi}{6} \)
(d) \( \frac{\pi}{2} \)
Answer: (d) \( \frac{\pi}{2} \)
In simple words: To find the principal argument of this complex fraction, it's best to convert both the numerator and denominator into polar form first. \( 1+i\sqrt{3} \) has modulus \( \sqrt{1^2+(\sqrt{3})^2}=2 \) and argument \( \tan^{-1}(\sqrt{3})=\frac{\pi}{3} \). So \( 1+i\sqrt{3} = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) \). Then \( (1+i\sqrt{3})^2 = 4(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}) \). For the denominator, \( 1-i\sqrt{3} = 2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) \). So \( 4i(1-i\sqrt{3}) = 4i \cdot 2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) = 8(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}) (\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) = 8(\cos(\frac{\pi}{2}-\frac{\pi}{3}) + i\sin(\frac{\pi}{2}-\frac{\pi}{3})) = 8(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}) \). Finally, divide the polar forms: \( \frac{4(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3})}{8(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6})} = \frac{1}{2}(\cos(\frac{2\pi}{3}-\frac{\pi}{6}) + i\sin(\frac{2\pi}{3}-\frac{\pi}{6})) = \frac{1}{2}(\cos(\frac{4\pi-\pi}{6}) + i\sin(\frac{4\pi-\pi}{6})) = \frac{1}{2}(\cos\frac{3\pi}{6} + i\sin\frac{3\pi}{6}) = \frac{1}{2}(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}) \). The principal argument is \( \frac{\pi}{2} \).

🎯 Exam Tip: When evaluating arguments of complex expressions involving multiplication, division, and powers, converting to polar form \( r(\cos\theta + i\sin\theta) \) and using De Moivre's theorem is often the most efficient method.

Question 21. If \( \alpha \) and \( \beta \) are the roots of \( x^2 + x + 1 = 0 \), then \( \alpha^{2020} + \beta^{2020} \) is:
(a) -2
(b) -1
(c) 1
(d) 2
Answer: (b) -1
In simple words: The equation \( x^2 + x + 1 = 0 \) has special roots. These roots are the non-real cube roots of unity, commonly denoted as \( \omega \) and \( \omega^2 \). So, \( \alpha = \omega \) and \( \beta = \omega^2 \) (or vice versa). We know that \( \omega^3 = 1 \). To find \( \omega^{2020} \), divide \( 2020 \) by \( 3 \): \( 2020 = 3 \times 673 + 1 \). So \( \omega^{2020} = (\omega^3)^{673} \cdot \omega^1 = 1^{673} \cdot \omega = \omega \). Similarly, \( (\omega^2)^{2020} = \omega^{4040} \). \( 4040 = 3 \times 1346 + 2 \), so \( \omega^{4040} = (\omega^3)^{1346} \cdot \omega^2 = 1^{1346} \cdot \omega^2 = \omega^2 \). Therefore, \( \alpha^{2020} + \beta^{2020} = \omega + \omega^2 \). Since \( 1 + \omega + \omega^2 = 0 \), we have \( \omega + \omega^2 = -1 \).

🎯 Exam Tip: Recognize that \( x^2+x+1=0 \) yields the non-real cube roots of unity. Always simplify powers of \( \omega \) by using the property \( \omega^3=1 \), reducing the exponent modulo 3.

Question 22. The product of all four values of \( \left(\cos\frac{\pi}{3} + i \sin \frac{\pi}{3}\right)^{\frac{3}{4}} \) is:
(a) -2
(b) -1
(c) 1
(d) 2
Answer: (c) 1
In simple words: When you have a complex number raised to a fractional power like \( (Z)^{\frac{p}{q}} \), it has \( q \) distinct values. In this case, \( q=4 \), so there are four values. Let the base be \( Z = \cos\frac{\pi}{3} + i \sin\frac{\pi}{3} = e^{i\frac{\pi}{3}} \). The four values of \( Z^{\frac{3}{4}} \) are \( e^{i\frac{3}{4}(\frac{\pi}{3} + 2k\pi)} \) for \( k=0, 1, 2, 3 \). The product of these \( q \) roots of a complex number is \( (-1)^{q-1} Z^p \). For this problem, \( q=4 \) (even), so the product is \( (-1)^{4-1} Z^{3} = -Z^3 \). Wait, actually for roots of unity, the product is usually 1 or -1. For \( q \)th roots of unity, product is \( (-1)^{q-1} \). For \( q \)th roots of an arbitrary complex number \( w \), say \( w^{1/q} \), the product is \( (-1)^{q-1} w \). So the product of \( w^{p/q} \) would be \( (-1)^{q-1} w^p \). Here \( w = e^{i\pi/3} \) and \( p=3, q=4 \). So the product is \( (-1)^{4-1} (e^{i\pi/3})^3 = (-1) (e^{i\pi}) = (-1) (-1) = 1 \).

🎯 Exam Tip: For any non-zero complex number \( z \), the product of its \( n \) distinct \( n \)th roots is \( (-1)^{n-1} z \). This general property saves time from calculating each root individually.

Question 23. If \( \omega \neq 1 \) is a cubic root of unity and \( \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix} = 3k \), then k is equal to:
(a) 1
(b) -1
(c) \( \sqrt{3}i \)
(d) \( -\sqrt{3}i \)
Answer: (d) \( -\sqrt{3}i \)
Given that \( \omega \) is a cubic root of unity, we know that \( \omega^3 = 1 \) and \( \omega^4 = \omega \). The problem asks us to find the value of k.
First, let's evaluate the given determinant expression by expanding it:
\( \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix} = 1(\omega^2 - \omega^4) - (\omega - \omega^2) + 1(\omega^2 - \omega) \)
Now, we substitute \( \omega^4 = \omega \) into the expression to simplify:
\( = (\omega^2 - \omega) - (\omega - \omega^2) + (\omega^2 - \omega) \)
\( = \omega^2 - \omega - \omega + \omega^2 + \omega^2 - \omega \)
\( = 3\omega^2 - 3\omega \)
We are given that this determinant is equal to \( 3k \). So, we can write:
\( 3k = 3\omega^2 - 3\omega \)
Divide both sides by 3 to find k:
\( k = \omega^2 - \omega \)
We know the standard values for \( \omega \) and \( \omega^2 \) for cubic roots of unity:
\( \omega = \frac{-1+\sqrt{3}i}{2} \)
\( \omega^2 = \frac{-1-\sqrt{3}i}{2} \)
Substitute these values into the expression for k:
\( k = \frac{-1-\sqrt{3}i}{2} - \left( \frac{-1+\sqrt{3}i}{2} \right) \)
\( k = \frac{-1-\sqrt{3}i + 1 - \sqrt{3}i}{2} \)
\( k = \frac{-2\sqrt{3}i}{2} \)
\( k = -\sqrt{3}i \)
In simple words: We used the special properties of cubic roots of unity, like \( \omega^3=1 \), to simplify the given determinant expression step-by-step. This allowed us to find a simpler form for 3k. After that, we used the actual complex values for \( \omega \) and \( \omega^2 \) to calculate the final value of k, which turned out to be an imaginary number.

🎯 Exam Tip: Always remember the fundamental properties of cubic roots of unity, such as \( \omega^3=1 \) and \( 1+\omega+\omega^2=0 \), as they simplify complex calculations significantly.

Question 24. The value of \( \left( \frac{1+\sqrt{3} i}{1-\sqrt{3} i} \right)^{10} \) is:
(a) \( \text{cis}\frac{2\pi}{3} \)
(b) \( \text{cis}\frac{4\pi}{3} \)
(c) \( -\text{cis}\frac{2\pi}{3} \)
(d) \( -\text{cis}\frac{2\pi}{3} \)
Answer: (a) \( \text{cis}\frac{2\pi}{3} \)
To find the value of the given expression, we first convert the complex numbers in the numerator and denominator into polar form.
For the numerator, \( 1+\sqrt{3}i \):
The modulus (distance from origin) is \( |1+\sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2 \).
The argument (angle) is \( \arg(1+\sqrt{3}i) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \).
So, \( 1+\sqrt{3}i = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) \).
For the denominator, \( 1-\sqrt{3}i \):
The modulus is \( |1-\sqrt{3}i| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2 \).
The argument is \( \arg(1-\sqrt{3}i) = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3} \).
So, \( 1-\sqrt{3}i = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right) \).
Now, let's divide the two complex numbers:
\( \frac{1+\sqrt{3}i}{1-\sqrt{3}i} = \frac{2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)}{2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)} \)
When dividing complex numbers in polar form, we divide their moduli and subtract their arguments:
\( = \text{cis}\left(\frac{\pi}{3} - \left(-\frac{\pi}{3}\right)\right) = \text{cis}\left(\frac{\pi}{3} + \frac{\pi}{3}\right) = \text{cis}\left(\frac{2\pi}{3}\right) \).
Finally, we need to raise this result to the power of 10 using De Moivre's Theorem:
\( \left( \text{cis}\frac{2\pi}{3} \right)^{10} = \text{cis}\left(10 \cdot \frac{2\pi}{3}\right) = \text{cis}\left(\frac{20\pi}{3}\right) \).
To find the principal argument, we subtract multiples of \( 2\pi \):
\( \frac{20\pi}{3} = 6\pi + \frac{2\pi}{3} = 3 \cdot 2\pi + \frac{2\pi}{3} \).
So, \( \text{cis}\left(\frac{20\pi}{3}\right) = \text{cis}\left(\frac{2\pi}{3}\right) \). This matches option (a).
In simple words: First, we change the complex numbers into a polar form, which uses a length and an angle. Then, we divide these polar forms by subtracting their angles. Lastly, we raise the result to the power of 10 using De Moivre's Theorem, which means multiplying the angle by 10. We then adjust the angle to its simplest form.

🎯 Exam Tip: When dealing with powers of complex numbers, always convert them to polar form first, as De Moivre's Theorem simplifies calculations significantly and helps avoid errors.

Question 25. If \( \omega = \text{cis}\frac{2\pi}{3} \), then the number of distinct roots of \( \begin{vmatrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = 0 \) is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1
Given the determinant equation, we need to find the number of distinct roots for z. Since \( \omega = \text{cis}\frac{2\pi}{3} \), it is a cubic root of unity, meaning \( \omega^3 = 1 \) and \( 1+\omega+\omega^2 = 0 \).
Let's expand the determinant along the first row:
\( \begin{vmatrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = (z+1)[(z+\omega^2)(z+\omega) - 1 \cdot 1] - \omega[\omega(z+\omega) - 1 \cdot \omega^2] + \omega^2[\omega \cdot 1 - \omega^2(z+\omega^2)] = 0 \)
Now, let's simplify each part of the expansion:
First term: \( (z+1)[(z^2 + z\omega + z\omega^2 + \omega^3) - 1] \)
\( = (z+1)[z^2 + z(\omega+\omega^2) + 1 - 1] \)
Since \( \omega+\omega^2 = -1 \) and \( \omega^3=1 \), this becomes:
\( = (z+1)[z^2 - z] \)
\( = z(z+1)(z-1) = z(z^2-1) = z^3 - z \)
Second term: \( -\omega[\omega z + \omega^2 - \omega^2] \)
\( = -\omega[z\omega] \)
\( = -z\omega^2 \)
Third term: \( +\omega^2[\omega - \omega^2 z - \omega^4] \)
Since \( \omega^4 = \omega \), this becomes:
\( = +\omega^2[\omega - \omega^2 z - \omega] \)
\( = +\omega^2[- \omega^2 z] \)
\( = -z\omega^4 = -z\omega \)
Now, substitute these simplified terms back into the determinant equation:
\( (z^3 - z) - z\omega^2 - z\omega = 0 \)
Factor out z from the terms with \( \omega \):
\( z^3 - z(1 + \omega^2 + \omega) = 0 \)
Since \( 1+\omega+\omega^2 = 0 \) for cubic roots of unity, we have:
\( z^3 - z(0) = 0 \)
\( z^3 = 0 \)
This equation has only one distinct root, which is \( z=0 \). Therefore, the number of distinct roots is 1.
In simple words: We used a method to break down the determinant (a special kind of number arrangement) into simpler parts. By using the properties of complex cubic roots of unity, we simplified these parts until the whole equation became \( z^3 = 0 \). This equation only has one unique solution, which is \( z=0 \).

🎯 Exam Tip: For determinants involving roots of unity, always try simplifying columns or rows using the property \( 1+\omega+\omega^2=0 \) or expanding carefully and simplifying terms using \( \omega^3=1 \).