Conjugate and Modulus of a Complex Number — JEE Main Maths PYQ MCQs with Solutions

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: $z = \dfrac{(1+i)^2}{a - i} = \dfrac{2i(a + i)}{a^2 + 1} = \dfrac{-2 + 2ai}{a^2 + 1}$, so $|z| = \dfrac{2}{\sqrt{a^2 + 1}}$. Setting $\dfrac{4}{a^2 + 1} = \dfrac{2}{5}$ gives $a^2 = 9$, so $a = 3$ (as $a > 0$). Then $z = \dfrac{-2 + 6i}{10} = -\dfrac{1}{5} + \dfrac{3}{5}i$, hence $\bar{z} = -\dfrac{1}{5} - \dfrac{3}{5}i$.

Solution: $|z_1| = 9$ is a circle with centre $C_1(0, 0)$ and radius $r_1 = 9$; $|z_2 - 3 - 4i| = 4$ is a circle with centre $C_2(3, 4)$ and radius $r_2 = 4$. Since $C_1C_2 = \sqrt{9 + 16} = 5 = r_1 - r_2$, the second circle touches the first internally, so the minimum distance between points of the two circles is $0$.

Solution: If $w = \dfrac{z - \alpha}{z + \alpha}$ is purely imaginary, then $w + \bar{w} = 0$, i.e. $\dfrac{z - \alpha}{z + \alpha} + \dfrac{\bar{z} - \alpha}{\bar{z} + \alpha} = 0$. Expanding the numerator gives $2z\bar{z} - 2\alpha^2 = 0$, so $\alpha^2 = |z|^2 = 4$ and $\alpha = \pm 2$. Hence a value of $\alpha$ is $2$.

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: $\left|\dfrac{z_1 - 2z_2}{2 - z_1\bar{z}_2}\right| = 1 \Rightarrow |z_1 - 2z_2|^2 = |2 - z_1\bar{z}_2|^2$. Using $|w|^2 = w\bar{w}$ and expanding, $|z_1|^2 + 4|z_2|^2 - 4 - |z_1|^2|z_2|^2 = 0$, i.e. $(|z_2|^2 - 1)(4 - |z_1|^2) = 0$. Since $|z_2| \neq 1$, we get $|z_1| = 2$: the point $z_1$ lies on a circle of radius $2$.

Solution: $|z| \geq 2$ is the region on or outside the circle of radius $2$ centred at the origin, and $\left|z + \dfrac{1}{2}\right|$ is the distance of $z$ from the point $-\dfrac{1}{2}$. The nearest point of the region is $(-2, 0)$, so the minimum value is $2 - \dfrac{1}{2} = \dfrac{3}{2}$, which lies in the interval $(1, 2)$.

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: Let $z = x + iy$ with $y \neq 0$. Then $a = (x^2 - y^2 + x + 1) + iy(2x + 1)$. For $a$ to be real, $y(2x + 1) = 0 \Rightarrow x = -\dfrac{1}{2}$. Then $a = \dfrac{1}{4} - y^2 - \dfrac{1}{2} + 1 = \dfrac{3}{4} - y^2 < \dfrac{3}{4}$ since $y \neq 0$. Hence $a$ cannot equal $\dfrac{3}{4}$.

Solution: $\bar{z}z^3 + z\bar{z}^3 = z\bar{z}(z^2 + \bar{z}^2) = 2(x^2 + y^2)(x^2 - y^2) = 350$, so $(x^2 + y^2)(x^2 - y^2) = 175$. For integer $x, y$ the only possibility is $x^2 + y^2 = 25$ and $x^2 - y^2 = 7$, giving $x = \pm 4$, $y = \pm 3$. The rectangle has sides $8$ and $6$, so its area is $8 \times 6 = 48$.

Solution: Since $|z| = 1$, $z\bar{z} = 1$, so $\dfrac{z}{1 - z^2} = \dfrac{z}{z\bar{z} - z^2} = \dfrac{1}{\bar{z} - z} = \dfrac{1}{-2i\,\text{Im}(z)}$, which is purely imaginary (as $z \neq \pm 1$ makes $\text{Im}(z) \neq 0$). Hence all such values lie on the $Y$-axis.

Solution: Let $u = \dfrac{w - \bar{w}z}{1 - z}$. Purely real means $u = \bar{u}$, i.e. $\dfrac{w - \bar{w}z}{1 - z} = \dfrac{\bar{w} - w\bar{z}}{1 - \bar{z}}$. Cross-multiplying and simplifying gives $(w - \bar{w})(1 - |z|^2) = 0$. Since $\beta \neq 0$, $w \neq \bar{w}$, so $|z|^2 = 1$. Hence $|z| = 1$ and $z \neq 1$.

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: Since $|z_k| = 1$, we have $z_k\bar{z}_k = 1$, so $\dfrac{1}{z_k} = \bar{z}_k$. Hence $1 = \left|\dfrac{1}{z_1} + \dfrac{1}{z_2} + \dfrac{1}{z_3}\right| = |\bar{z}_1 + \bar{z}_2 + \bar{z}_3| = \left|\overline{z_1 + z_2 + z_3}\right| = |z_1 + z_2 + z_3|$.

Solution: Since $i^3 = i^7 = -i$ and $i^5 = i$, the expression equals $\left[(1 + i)^{n_1} + (1 - i)^{n_1}\right] + \left[(1 + i)^{n_2} + (1 - i)^{n_2}\right] = 2\,\text{Re}\,(1 + i)^{n_1} + 2\,\text{Re}\,(1 + i)^{n_2}$, because $w + \bar{w} = 2\,\text{Re}(w)$. This is real for every choice of positive integers, so the only condition needed is $n_1 > 0,\ n_2 > 0$.

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: $|w| = 1 \Rightarrow |1 - iz| = |z - i|$. Since $1 - iz = -i(z + i)$ and $|-i| = 1$, this gives $|z + i| = |z - i|$: $z$ is equidistant from $i$ and $-i$. The locus is the perpendicular bisector of $(0, 1)$ and $(0, -1)$, which is the real axis.

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: 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: $\dfrac{z_1 + z_2}{z_1 - z_2} = \dfrac{(z_1 + z_2)(\bar{z}_1 - \bar{z}_2)}{|z_1 - z_2|^2} = \dfrac{|z_1|^2 - |z_2|^2 + \bar{z}_1 z_2 - z_1\bar{z}_2}{|z_1 - z_2|^2} = \dfrac{2i\,\text{Im}(\bar{z}_1 z_2)}{|z_1 - z_2|^2}$, using $|z_1| = |z_2|$ and $w - \bar{w} = 2i\,\text{Im}(w)$. So the expression is always purely imaginary or zero; it can never be a non-zero real number.

Solution: Given $a^2 + b^2 = c^2 + d^2 = 1$ and $\text{Re}(z_1\bar{z}_2) = ac + bd = 0$, so $\dfrac{a}{b} = -\dfrac{d}{c} = \lambda$ (say), i.e. $a = \lambda b$, $d = -\lambda c$. Substituting into the modulus conditions gives $b^2 = c^2$ and $a^2 = d^2$. Then $|w_1|^2 = a^2 + c^2 = a^2 + b^2 = 1$, $|w_2|^2 = b^2 + d^2 = b^2 + a^2 = 1$, and $\text{Re}(w_1\bar{w}_2) = ab + cd = \lambda b^2 - \lambda c^2 = 0$.

Solution: $|2z - 6 + 5i| = 2\left|z - 3 + \dfrac{5i}{2}\right| = 2\left|(z - 3 - 2i) + \dfrac{9i}{2}\right| \geq 2\left(\dfrac{9}{2} - |z - 3 - 2i|\right) \geq 2\left(\dfrac{9}{2} - 2\right) = 5$, by the triangle inequality $|w_1 + w_2| \geq |w_2| - |w_1|$. The bound is attained on the boundary of the disc, so the minimum value is $5$.