Rotation of a Complex Number — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ Rotation of a Complex Number MCQs with step-by-step solutions (8 questions). Part of Complex Numbers. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Rotation of a Complex Number · easy · numerical
Let $z = \left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5 + \left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$, then
A. $R(z)>0$ and $I(z)>0$
B. $I(z)=0$ ✓ Correct
C. $R(z)<0$ and $I(z)>0$
D. $R(z)=-3$
Solution: In Euler form, $\frac{\sqrt{3}}{2}+\frac{i}{2}=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}=e^{i\pi/6}$ and $\frac{\sqrt{3}}{2}-\frac{i}{2}=e^{-i\pi/6}$. So $z=e^{i5\pi/6}+e^{-i5\pi/6}=2\cos\frac{5\pi}{6}=-\sqrt{3}$. Hence $z$ is purely real: $I(z)=0$ and $R(z)=-\sqrt{3}<0$ (not $-3$).
Q2 — Rotation of a Complex Number · medium · numerical
A particle $P$ starts from the point $z_0 = 1 + 2i$, where $i=\sqrt{-1}$. It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point $z_1$. From $z_1$ the particle moves $\sqrt{2}$ units in the direction of the vector $\hat{i}+\hat{j}$ and then it moves through an angle $\frac{\pi}{2}$ in anti-clockwise direction on a circle with centre at origin, to reach a point $z_2$. The point $z_2$ is given by
A. $6+7i$
B. $-7+6i$
C. $7+6i$
D. $-6+7i$ ✓ Correct
Solution: From $z_0=1+2i$, moving 5 units right and then 3 units up gives $z_1=6+5i$. Moving $\sqrt{2}$ units along $\hat{i}+\hat{j}$ adds $\sqrt{2}\left(\frac{1+i}{\sqrt{2}}\right)=1+i$, reaching $z_2'=7+6i$. Rotating anticlockwise by $\frac{\pi}{2}$ about the origin multiplies by $e^{i\pi/2}=i$, so $z_2=i(7+6i)=-6+7i$.
Q3 — Rotation of a Complex Number · medium · numerical
A man walks a distance of 3 units from the origin towards the North-East (N 45° E) direction. From there, he walks a distance of 4 units towards the North-West (N 45° W) direction to reach a point $P$. Then, the position of $P$ in the Argand plane is
A. $3e^{i\pi/4}+4i$
B. $(3-4i)\,e^{i\pi/4}$
C. $(4+3i)\,e^{i\pi/4}$
D. $(3+4i)\,e^{i\pi/4}$ ✓ Correct
Solution: After the first walk the man is at $A=3e^{i\pi/4}$. The North-West direction is the North-East direction rotated $90°$ anticlockwise, so the second displacement is $4\,e^{i\pi/4}\cdot i=4i\,e^{i\pi/4}$. Hence $P=3e^{i\pi/4}+4i\,e^{i\pi/4}=(3+4i)\,e^{i\pi/4}$.
Q4 — Rotation of a Complex Number · medium · theory
Let $0<\alpha<\frac{\pi}{2}$ be a fixed angle. If $P=(\cos\theta,\ \sin\theta)$ and $Q=\{\cos(\alpha-\theta),\ \sin(\alpha-\theta)\}$, then $Q$ is obtained from $P$ by
A. clockwise rotation around origin through an angle $\alpha$
B. anti-clockwise rotation around origin through an angle $\alpha$
C. reflection in the line through origin with slope $\tan\alpha$
D. reflection in the line through origin with slope $\tan\frac{\alpha}{2}$ ✓ Correct
Solution: On the Argand plane $P$ is $e^{i\theta}$ and $Q$ is $e^{i(\alpha-\theta)}$. A rotation would send $e^{i\theta}$ to $e^{i(\theta\pm\alpha)}$, which is not the case here. The map $\theta\mapsto\alpha-\theta$ fixes the direction $\theta=\frac{\alpha}{2}$ and reverses angles about it, which is exactly reflection in the line through the origin making angle $\frac{\alpha}{2}$ with the $X$-axis, i.e. the line of slope $\tan\frac{\alpha}{2}$.
Q5 — Rotation of a Complex Number · easy · numerical
The complex numbers $z_1, z_2$ and $z_3$ satisfying $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$ are the vertices of a triangle which is
A. of area zero
B. right angled isosceles
C. equilateral ✓ Correct
D. obtuse angled isosceles
Solution: $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}=\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}=e^{-i\pi/3}$. Its modulus is 1, so $|z_1-z_3|=|z_2-z_3|$, and its argument shows the angle at $z_3$ is $\frac{\pi}{3}$. Two equal sides enclosing $60°$ make the triangle equilateral.
Q6 — Rotation of a Complex Number · hard · numerical
Let $a, b \in R$ and $a^2+b^2 \neq 0$. Suppose $S=\left\{z \in C : z=\frac{1}{a+ibt},\ t \in R,\ t \neq 0\right\}$, where $i=\sqrt{-1}$. If $z=x+iy$ and $z \in S$, then $(x, y)$ lies on
A. the circle with radius $\frac{1}{2a}$ and centre $\left(\frac{1}{2a},\ 0\right)$ for $a>0,\ b \neq 0$ ✓ Correct
B. the circle with radius $-\frac{1}{2a}$ and centre $\left(-\frac{1}{2a},\ 0\right)$ for $a<0,\ b \neq 0$
C. the $X$-axis for $a \neq 0,\ b = 0$ ✓ Correct
D. the $Y$-axis for $a = 0,\ b \neq 0$ ✓ Correct
Solution: Rationalising, $x+iy=\frac{a-ibt}{a^2+b^2t^2}$, so $x=\frac{a}{a^2+b^2t^2}$ and $y=\frac{-bt}{a^2+b^2t^2}$. For $a\neq 0,\ b\neq 0$, eliminating $t$ gives $x^2+y^2=\frac{x}{a}$, i.e. $\left(x-\frac{1}{2a}\right)^2+y^2=\frac{1}{4a^2}$ — a circle with centre $\left(\frac{1}{2a},0\right)$ and radius $\frac{1}{2|a|}$, matching (a) for $a>0$; option (b) states the wrong centre $\left(-\frac{1}{2a},0\right)$. For $b=0$, $z=\frac{1}{a}$ lies on the $X$-axis; for $a=0$, $z=-\frac{i}{bt}$ lies on the $Y$-axis. Hence (a), (c), (d).
Q7 — Rotation of a Complex Number · medium · numerical
Let $w=\frac{\sqrt{3}+i}{2}$ and $P=\{w^n : n=1, 2, 3, \ldots\}$. Further $H_1=\left\{z \in C : \operatorname{Re}(z)>\frac{1}{2}\right\}$ and $H_2=\left\{z \in C : \operatorname{Re}(z)<\frac{-1}{2}\right\}$, where $C$ is the set of all complex numbers. If $z_1 \in P \cap H_1$, $z_2 \in P \cap H_2$ and $O$ represents the origin, then $\angle z_1 O z_2$ is equal to
A. $\frac{\pi}{2}$
B. $\frac{\pi}{6}$
C. $\frac{2\pi}{3}$ ✓ Correct
D. $\frac{5\pi}{6}$ ✓ Correct
Solution: $w=e^{i\pi/6}$, so $w^n=\cos\frac{n\pi}{6}+i\sin\frac{n\pi}{6}$ — points on the unit circle. $z_1\in P\cap H_1$ requires $\cos\frac{n\pi}{6}>\frac{1}{2}$, giving $z_1=\frac{\sqrt{3}}{2}\pm\frac{i}{2}$; $z_2\in P\cap H_2$ requires $\cos\frac{n\pi}{6}<-\frac{1}{2}$, giving $z_2=-\frac{\sqrt{3}}{2}\pm\frac{i}{2}$ or $-1$. The possible angles $\angle z_1Oz_2$ are $\frac{2\pi}{3},\ \frac{5\pi}{6}$ (and $\pi$), so both $\frac{2\pi}{3}$ and $\frac{5\pi}{6}$ are attainable.
Q8 — Rotation of a Complex Number · medium · numerical
For any integer $k$, let $\alpha_k=\cos\left(\frac{k\pi}{7}\right)+i\sin\left(\frac{k\pi}{7}\right)$, where $i=\sqrt{-1}$. The value of the expression $\frac{\sum_{k=1}^{12}|\alpha_{k+1}-\alpha_k|}{\sum_{k=1}^{3}|\alpha_{4k-1}-\alpha_{4k-2}|}$ is
Solution: Each $\alpha_k=e^{ik\pi/7}$ is a vertex of a regular 14-sided polygon inscribed in the unit circle, so every $|\alpha_{k+1}-\alpha_k|$ equals the same side length $a$. In the denominator the indices $4k-1$ and $4k-2$ also differ by 1, so each term is again $a$. The ratio is $\frac{12a}{3a}=4$.