Complex Numbers — JEE Main Maths PYQ MCQs with Solutions

Solution: Let $x + iy = \frac{\alpha + i}{\alpha - i} = \frac{(\alpha + i)^2}{\alpha^2 + 1} = \frac{\alpha^2 - 1}{\alpha^2 + 1} + \frac{2\alpha}{\alpha^2 + 1}i$. Then $x^2 + y^2 = \frac{(\alpha^2 - 1)^2 + 4\alpha^2}{(\alpha^2 + 1)^2} = \frac{(\alpha^2 + 1)^2}{(\alpha^2 + 1)^2} = 1$. So every point of $S$ lies on the circle $x^2 + y^2 = 1$, a circle of radius 1. (Directly: $|\alpha + i| = |\alpha - i|$ for real $\alpha$, so each element has modulus 1.)

Solution: Rationalising, $\frac{2 + 3i\sin\theta}{1 - 2i\sin\theta} = \frac{2 - 6\sin^2\theta}{1 + 4\sin^2\theta} + \frac{7\sin\theta}{1 + 4\sin^2\theta}i$. For a purely imaginary number the real part is zero: $2 - 6\sin^2\theta = 0 \Rightarrow \sin^2\theta = \frac{1}{3} \Rightarrow \sin\theta = \pm\frac{1}{\sqrt{3}}$. Hence $\theta = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$ is such a value.

Solution: $\sum_{n=1}^{13}(i^n + i^{n+1}) = (1 + i)\sum_{n=1}^{13} i^n$. The sum of any four consecutive powers of $i$ is zero, so the first 12 terms of $\sum i^n$ cancel, leaving $i^{13} = i$. Hence the sum is $(1+i)i = i + i^2 = i - 1$.

Solution: $\frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{2i}{2} = i$. So we need the smallest positive integer $n$ with $i^n = 1$, which is $n = 4$. Since 4 is not among the listed values, the answer is None of these.

Solution: Let $z = x + iy$. Then $|z - i| = |z - 1|$ gives $x^2 + (y-1)^2 = (x-1)^2 + y^2$, which simplifies to $y = x$ — the perpendicular bisector of the points $i$ and $1$. This is a line through the origin with slope $1$.

Solution: Let $z = x + iy$. Then $\sqrt{x^2 + y^2} + x + iy = 3 + i$ gives $y = 1$ and $\sqrt{x^2 + 1} = 3 - x$. Squaring: $x^2 + 1 = 9 - 6x + x^2 \Rightarrow x = \dfrac{4}{3}$. Hence $|z| = \sqrt{\dfrac{16}{9} + 1} = \dfrac{5}{3}$.

Solution: $2\,\text{Re}(w) = w + \bar{w} = \dfrac{z - 1}{z + 1} + \dfrac{\bar{z} - 1}{\bar{z} + 1} = \dfrac{(z - 1)(\bar{z} + 1) + (\bar{z} - 1)(z + 1)}{|z + 1|^2} = \dfrac{2z\bar{z} - 2}{|z + 1|^2} = \dfrac{2(|z|^2 - 1)}{|z + 1|^2} = 0$ since $|z| = 1$. Hence $\text{Re}(w) = 0$.

Solution: $z_1$ lies on the circle of radius $12$ centred at the origin, and $z_2$ on the circle of radius $5$ centred at $(3, 4)$, which is at distance $5$ from the origin. By the triangle inequality, $|z_1 - z_2| \geq |z_1| - |z_2 - (3 + 4i)| - |3 + 4i| = 12 - 5 - 5 = 2$, and equality is attained when $z_1, z_2$ lie along the common diameter. Hence the minimum value is $2$.

Solution: Conjugacy requires $\overline{\sin x + i\cos 2x} = \cos x - i\sin 2x$, i.e. $\sin x = \cos x$ and $\cos 2x = \sin 2x$, so $\tan x = 1$ and $\tan 2x = 1$. The first gives $x = \dfrac{\pi}{4} + n\pi$ and the second $x = \dfrac{\pi}{8} + \dfrac{n\pi}{2}$; these can never hold simultaneously, so no value of $x$ works.

Solution: In a parallelogram the diagonals bisect each other, so the midpoint of the diagonal $z_1z_3$ coincides with the midpoint of the diagonal $z_2z_4$: $\dfrac{z_1 + z_3}{2} = \dfrac{z_2 + z_4}{2}$, i.e. $z_1 + z_3 = z_2 + z_4$.

Solution: $|z - 4| < |z - 2|$ means $z$ is closer to $4$ than to $2$, i.e. $z$ lies on the right of the perpendicular bisector of the points $2$ and $4$. Writing $z = x + iy$ and squaring gives $x > 3$, i.e. $\text{Re}(z) > 3$, which is none of the listed regions.

Solution: The two terms are conjugates, and $w^5 + \bar{w}^5 = w^5 + \overline{w^5} = 2\,\text{Re}(w^5)$ is always real, so $\text{Im}(z) = 0$. In fact $\dfrac{\sqrt{3}}{2} + \dfrac{i}{2} = e^{i\pi/6}$, so $z = 2\cos\dfrac{5\pi}{6} = -\sqrt{3}$ (and $\text{Re}(z) < 0$).

Solution: $\left|\dfrac{z - 5i}{z + 5i}\right| = 1 \Rightarrow |z - 5i| = |z + 5i|$, so $z$ is equidistant from $(0, 5)$ and $(0, -5)$. The locus is the perpendicular bisector of these two points, which is the $X$-axis.

Solution: Since $|z| = 1$, we have $\bar{z} = \frac{1}{z}$. Therefore $\frac{1+z}{1+\bar{z}} = \frac{1+z}{1+\frac{1}{z}} = \frac{z(1+z)}{z+1} = z$. Hence $\arg\left(\frac{1+z}{1+\bar{z}}\right) = \arg(z) = \theta$.

Solution: Let $w = re^{i\theta}$. Then $|z| = r$ and $\arg(z) = \pi - \theta$, so $z = re^{i(\pi-\theta)} = re^{i\pi}e^{-i\theta} = -re^{-i\theta} = -\bar{w}$.

Solution: Squaring both sides gives $|z_1|^2 + |z_2|^2 + 2|z_1||z_2|\cos(\arg z_1 - \arg z_2) = |z_1|^2 + |z_2|^2 + 2|z_1||z_2|$, so $\cos(\arg z_1 - \arg z_2) = 1$, i.e. $\arg(z_1) - \arg(z_2) = 0$. Equality in the triangle inequality holds exactly when $z_1$ and $z_2$ point in the same direction.

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 $-3$).

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.

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 $\arg z = \frac{\pi}{4}$.

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 $n = 3$.

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 $4 + 5\omega + 3\omega^2 = 1 + 3(1 + \omega + \omega^2) + 2\omega = 1 + 2\omega = i\sqrt{3}$.

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 $\omega^{12} = 1$.

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 $B = 1$.

Solution: $\frac{az+b}{z+1} = \frac{(ax+b) + iay}{(x+1) + iy}$, so $\text{Im}\left(\frac{az+b}{z+1}\right) = \frac{ay(x+1) - (ax+b)y}{(x+1)^2 + y^2} = \frac{y(a-b)}{(x+1)^2 + y^2} = \frac{y}{(x+1)^2 + y^2}$ using $a - b = 1$. Setting this equal to $y$ and cancelling $y \neq 0$ gives $(x+1)^2 + y^2 = 1$, so $x = -1 \pm \sqrt{1 - y^2}$.

Solution: The conditions give $|\alpha - z_0|^2 = r^2$ and $\left|\dfrac{1}{\bar{\alpha}} - z_0\right|^2 = 4r^2$. Expanding: $|\alpha|^2 - (\bar{\alpha}z_0 + \alpha\bar{z}_0) + |z_0|^2 = r^2$, and multiplying the second through by $|\alpha|^2$: $1 - (\bar{\alpha}z_0 + \alpha\bar{z}_0) + |\alpha|^2|z_0|^2 = 4r^2|\alpha|^2$. Subtracting, $(1 - |\alpha|^2)(1 - |z_0|^2) = r^2(4|\alpha|^2 - 1)$. With $|z_0|^2 = \dfrac{r^2 + 2}{2}$ this becomes $\dfrac{|\alpha|^2 - 1}{2} = 4|\alpha|^2 - 1$, so $7|\alpha|^2 = 1$ and $|\alpha| = \dfrac{1}{\sqrt{7}}$.

Solution: Taking the conjugate of $sz + t\bar{z} + r = 0$ gives $\bar{s}\bar{z} + \bar{t}z + \bar{r} = 0$; eliminating $\bar{z}$ yields $(|s|^2 - |t|^2)z = \bar{r}t - r\bar{s}$. If $L$ has exactly one element then $|s|^2 - |t|^2 \neq 0$, so (a) is true. If $|s| = |t|$, then $\bar{r}t - r\bar{s}$ may or may not vanish, so $L$ can be empty — (b) is false; when solutions do exist, $L$ is a line, so more than one element forces infinitely many — (d) is true. A line (or a single point) meets the circle $|z - 1 + i| = 5$ in at most two points, so (c) is true.

Solution: (a) $-1-i$ lies in the third quadrant, so $\arg(-1-i) = -\frac{3\pi}{4} \ne \frac{\pi}{4}$ — FALSE. (b) As $t \to 0^-$, $\arg(-1+it) \to -\pi$ while $f(0) = \arg(-1) = \pi$, so $f$ is discontinuous at $t = 0$ — FALSE. (c) $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2) + 2n\pi$ for some integer $n$, so the expression equals $2n\pi$ — TRUE. (d) The condition makes the cross-ratio $\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$ purely real (and negative), which means $z$ lies on the circle through $z_1, z_2, z_3$, not on a straight line — FALSE. Hence (a), (b), (d) are the false statements.

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).

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 $3$.

Solution: Let $z = x + 10i$. Then $2z - n = (2i-1)(2z+n)$ gives $(2x - n) + 20i = -(2x + n + 40) + (4x + 2n - 20)i$. Comparing real parts: $2x - n = -2x - n - 40 \Rightarrow 4x = -40 \Rightarrow x = -10$. Comparing imaginary parts: $20 = 4x + 2n - 20 \Rightarrow 20 = -40 + 2n - 20 \Rightarrow n = 40$. So $n = 40$ and $\text{Re}(z) = -10$.