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De-Moivre's Theorem, Cube Roots and nth Roots of Unity — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ De-Moivre's Theorem, Cube Roots and nth Roots of Unity MCQs with step-by-step solutions (13 questions). Part of Complex Numbers. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · medium · theory
If $z$ and $w$ are two complex numbers such that $|zw| = 1$ and $\arg(z) - \arg(w) = \frac{\pi}{2}$, then
A. $\bar{z}w = -i$ ✓ Correct
B. $\bar{z}w = \frac{1-i}{\sqrt{2}}$
C. $\bar{z}w = i$
D. $\bar{z}w = \frac{-1+i}{\sqrt{2}}$
Solution: Let $|z| = r$; then $|zw| = 1$ gives $|w| = \frac{1}{r}$. Let $\arg(w) = \theta$, so $\arg(z) = \frac{\pi}{2} + \theta$. Then $z = re^{i(\pi/2 + \theta)}$ and $w = \frac{1}{r}e^{i\theta}$, so $\bar{z}w = re^{-i(\pi/2 + \theta)} \cdot \frac{1}{r}e^{i\theta} = e^{-i\pi/2} = -i$.
Q2 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · medium · numerical
If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$ $\left(i = \sqrt{-1}\right)$, then $\left(1 + iz + z^5 + iz^8\right)^9$ is equal to
A. $1$
B. $(-1 + 2i)^9$
C. $-1$ ✓ Correct
D. $0$
Solution: $z = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = e^{i\pi/6}$. Then $iz = e^{i2\pi/3}$, $z^5 = e^{i5\pi/6}$ and $iz^8 = e^{i11\pi/6}$, so $1 + iz + z^5 + iz^8 = 1 + \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) + \left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) + \left(\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = \frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i\pi/3}$. Hence the expression equals $\left(e^{i\pi/3}\right)^9 = e^{i3\pi} = -1$.
Q3 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · easy · numerical
Let $z_0$ be a root of the quadratic equation $x^2 + x + 1 = 0$. If $z = 3 + 6iz_0^{81} - 3iz_0^{93}$, then $\arg z$ is equal to
A. $\frac{\pi}{4}$ ✓ Correct
B. $\frac{\pi}{6}$
C. $0$
D. $\frac{\pi}{3}$
Solution: The roots of $x^2 + x + 1 = 0$ are the non-real cube roots of unity, so $z_0^3 = 1$. Since $81$ and $93$ are multiples of $3$, $z_0^{81} = z_0^{93} = 1$. Thus $z = 3 + 6i - 3i = 3 + 3i$, which lies in the first quadrant with $\tan\theta = \frac{3}{3} = 1$, so $\arg z = \frac{\pi}{4}$.
Q4 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · medium · numerical
Let $z = \cos\theta + i\sin\theta$. Then the value of $\sum_{m=1}^{15} \operatorname{Im}\left(z^{2m-1}\right)$ at $\theta = 2^\circ$ is
A. $\frac{1}{\sin 2^\circ}$
B. $\frac{1}{3\sin 2^\circ}$
C. $\frac{1}{2\sin 2^\circ}$
D. $\frac{1}{4\sin 2^\circ}$ ✓ Correct
Solution: By De Moivre's theorem $\operatorname{Im}(z^{2m-1}) = \sin(2m-1)\theta$, so the sum is $\sin\theta + \sin 3\theta + \cdots + \sin 29\theta$. Using the sum of sines in AP, this equals $\frac{\sin(15\theta)}{\sin\theta}\cdot\sin(15\theta) = \frac{\sin^2 15\theta}{\sin\theta}$. At $\theta = 2^\circ$ it becomes $\frac{\sin^2 30^\circ}{\sin 2^\circ} = \frac{1}{4\sin 2^\circ}$.
Q5 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · medium · theory
The minimum value of $|a + b\omega + c\omega^2|$, where $a$, $b$ and $c$ are all not equal integers and $\omega\,(\neq 1)$ is a cube root of unity, is
A. $\sqrt{3}$
B. $\frac{1}{2}$
C. $1$ ✓ Correct
D. $0$
Solution: $|a + b\omega + c\omega^2|^2 = a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}\left[(a-b)^2 + (b-c)^2 + (c-a)^2\right]$. Since $a, b, c$ are integers not all equal, at most one of the squared differences can be $0$ and the others are $\geq 1$, so the square is at least $\frac{1}{2}(0 + 1 + 1) = 1$. Hence the minimum value of $|a + b\omega + c\omega^2|$ is $1$.
Q6 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · easy · numerical
If $\omega\,(\neq 1)$ be a cube root of unity and $(1 + \omega^2)^n = (1 + \omega^4)^n$, then the least positive value of $n$ is
A. $2$
B. $3$ ✓ Correct
C. $5$
D. $6$
Solution: Using $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$: $1 + \omega^2 = -\omega$ and $1 + \omega^4 = 1 + \omega = -\omega^2$. So $(-\omega)^n = (-\omega^2)^n \Rightarrow \omega^n = \omega^{2n} \Rightarrow \omega^n = 1$, whose least positive solution is $n = 3$.
Q7 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · medium · numerical
Let $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Then the value of the determinant $\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4 \end{vmatrix}$ is
A. $3\omega$
B. $3\omega(\omega - 1)$ ✓ Correct
C. $3\omega^2$
D. $3\omega(1 - \omega)$
Solution: Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$, then expand along the first column: the determinant equals $(-2 - \omega^2)(\omega^4 - 1) - (\omega^2 - 1)^2$. Using $\omega^3 = 1$ (so $\omega^4 = \omega$) and $1 + \omega + \omega^2 = 0$, this simplifies to $3\omega^2 - 3\omega = 3\omega(\omega - 1)$.
Q8 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · medium · theory
Let $z_1$ and $z_2$ be $n$th roots of unity which subtend a right angle at the origin. Then $n$ must be of the form (where $k$ is an integer)
A. $4k + 1$
B. $4k + 2$
C. $4k + 3$
D. $4k$ ✓ Correct
Solution: Since $z_1$ and $z_2$ subtend a right angle at the origin, $\arg\frac{z_1}{z_2} = \pm\frac{\pi}{2}$, and as $|z_1| = |z_2| = 1$ we get $\frac{z_1}{z_2} = \pm i$. Both being $n$th roots of unity, $\left(\frac{z_1}{z_2}\right)^n = 1$, i.e. $(\pm i)^n = 1$, which requires $n$ to be a multiple of $4$. Hence $n = 4k$.
Q9 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · easy · numerical
If $i = \sqrt{-1}$, then $4 + 5\left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)^{334} + 3\left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)^{365}$ is equal to
A. $1 - i\sqrt{3}$
B. $-1 + i\sqrt{3}$
C. $i\sqrt{3}$ ✓ Correct
D. $-i\sqrt{3}$
Solution: Here $-\frac{1}{2} + \frac{i\sqrt{3}}{2} = \omega$, a cube root of unity. Since $334 = 3(111) + 1$ and $365 = 3(121) + 2$, we have $\omega^{334} = \omega$ and $\omega^{365} = \omega^2$. So the expression is $4 + 5\omega + 3\omega^2 = 1 + 3(1 + \omega + \omega^2) + 2\omega = 1 + 2\omega = i\sqrt{3}$.
Q10 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · easy · numerical
If $\omega$ is an imaginary cube root of unity, then $(1 + \omega - \omega^2)^7$ is equal to
A. $128\omega$
B. $-128\omega$
C. $128\omega^2$
D. $-128\omega^2$ ✓ Correct
Solution: Since $1 + \omega + \omega^2 = 0$, we have $1 + \omega = -\omega^2$, so $1 + \omega - \omega^2 = -2\omega^2$. Therefore $(1 + \omega - \omega^2)^7 = (-2\omega^2)^7 = -128\,\omega^{14} = -128\,\omega^2$, using $\omega^{12} = 1$.
Q11 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · easy · numerical
If $\omega\,(\neq 1)$ is a cube root of unity and $(1 + \omega)^7 = A + B\omega$, then $A$ and $B$ are respectively
A. $0,\ 1$
B. $1,\ 1$ ✓ Correct
C. $1,\ 0$
D. $-1,\ 1$
Solution: $(1 + \omega)^7 = (1 + \omega)(1 + \omega)^6 = (1 + \omega)(-\omega^2)^6 = (1 + \omega)\,\omega^{12} = 1 + \omega$, since $1 + \omega = -\omega^2$ and $\omega^{12} = 1$. Comparing with $A + B\omega$ gives $A = 1$ and $B = 1$.
Q12 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · medium · numerical
The value of $\sum_{k=1}^{6}\left(\sin\frac{2\pi k}{7} - i\cos\frac{2\pi k}{7}\right)$ is
A. $-1$
B. $0$
C. $-i$
D. $i$ ✓ Correct
Solution: The sum equals $-i\sum_{k=1}^{6}\left(\cos\frac{2\pi k}{7} + i\sin\frac{2\pi k}{7}\right) = -i\sum_{k=1}^{6} e^{i2\pi k/7}$. The seven $7$th roots of unity sum to $0$, so $\sum_{k=1}^{6} e^{i2\pi k/7} = -1$. Hence the value is $-i(-1) = i$.
Q13 — De-Moivre's Theorem, Cube Roots and nth Roots of Unity · hard · numerical
Let $\omega = e^{i\pi/3}$, and $a, b, c, x, y, z$ be non-zero complex numbers such that $a + b + c = x$, $a + b\omega + c\omega^2 = y$, $a + b\omega^2 + c\omega = z$. Then the value of $\frac{|x|^2 + |y|^2 + |z|^2}{|a|^2 + |b|^2 + |c|^2}$ is
Solution: As noted in the book's solution, the printed $\omega = e^{i\pi/3}$ is a misprint for $\omega = e^{i2\pi/3}$, a cube root of unity — only then does the ratio take a fixed integer value. With $\omega = e^{i2\pi/3}$, expand $|x|^2 + |y|^2 + |z|^2 = x\bar{x} + y\bar{y} + z\bar{z}$; using $1 + \omega + \omega^2 = 0$ and $\bar{\omega} = \omega^2$, all cross terms cancel and the sum equals $3\left(|a|^2 + |b|^2 + |c|^2\right)$. Hence the required value is $3$.