Argument of a Complex Number — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ Argument 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 — Argument of a Complex Number · easy · theory
If $z$ is a complex number of unit modulus and argument $\theta$, then $\arg\left(\frac{1+z}{1+\bar{z}}\right)$ is equal to
A. $-\theta$
B. $\frac{\pi}{2}-\theta$
C. $\theta$ ✓ Correct
D. $\pi-\theta$
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 $\arg\left(\frac{1+z}{1+\bar{z}}\right) = \arg(z) = \theta$.
Q2 — Argument of a Complex Number · medium · theory
If $\arg(z) < 0$, then $\arg(-z) - \arg(z)$ equals
A. $\pi$ ✓ Correct
B. $-\pi$
C. $-\frac{\pi}{2}$
D. $\frac{\pi}{2}$
Solution: Let $\arg(z) = -\theta$ where $0 < \theta < \pi$. Then $-z$ makes an angle $\pi - \theta$ with the positive real axis, and since $0 < \pi - \theta < \pi$, this is already the principal value: $\arg(-z) = \pi - \theta$. Hence $\arg(-z) - \arg(z) = (\pi - \theta) - (-\theta) = \pi$.
Q3 — Argument of a Complex Number · medium · theory
Let $z$ and $w$ be two complex numbers such that $|z| \le 1$, $|w| \le 1$ and $|z + iw| = |z - i\bar{w}| = 2$. Then $z$ equals
A. $1$ or $i$
B. $i$ or $-i$
C. $1$ or $-1$ ✓ Correct
D. $i$ or $-1$
Solution: By the triangle inequality, $|z + iw| \le |z| + |w| \le 2$; equality forces $|z| = |w| = 1$ with $z$ and $iw$ having the same argument, i.e. $z = iw$. Similarly $|z - i\bar{w}| = 2$ forces $z = -i\bar{w}$. Multiplying the two, $z^2 = (iw)(-i\bar{w}) = w\bar{w} = |w|^2 = 1$, so $z = 1$ or $z = -1$.
Q4 — Argument of a Complex Number · easy · theory
Let $z$ and $w$ be two non-zero complex numbers such that $|z| = |w|$ and $\arg(z) + \arg(w) = \pi$. Then $z$ equals
A. $w$
B. $-w$
C. $\bar{w}$
D. $-\bar{w}$ ✓ Correct
Solution: Let $w = re^{i\theta}$. Then $|z| = r$ and $\arg(z) = \pi - \theta$, so $z = re^{i(\pi-\theta)} = re^{i\pi}e^{-i\theta} = -re^{-i\theta} = -\bar{w}$.
Q5 — Argument of a Complex Number · easy · theory
If $z_1$ and $z_2$ are two non-zero complex numbers such that $|z_1 + z_2| = |z_1| + |z_2|$, then $\arg(z_1) - \arg(z_2)$ is equal to
A. $-\pi$
B. $-\frac{\pi}{2}$
C. $0$ ✓ Correct
D. $\frac{\pi}{2}$
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.
Q6 — Argument of a Complex Number · medium · theory
If $a, b, c$ and $u, v, w$ are the complex numbers representing the vertices of two triangles such that $c = (1-r)a + rb$ and $w = (1-r)u + rv$, where $r$ is a complex number, then the two triangles
A. have the same area
B. are similar ✓ Correct
C. are congruent
D. None of these
Solution: The relations give $c - a = r(b - a)$ and $w - u = r(v - u)$, so $\frac{c-a}{b-a} = \frac{w-u}{v-u} = r$. Equal complex ratios mean the corresponding sides are in the same proportion $|r|$ with the same included angle $\arg(r)$, hence the triangles are similar. Equivalently, the similarity determinant $\begin{vmatrix} a & u & 1 \\ b & v & 1 \\ c & w & 1 \end{vmatrix}$ vanishes.
Q7 — Argument of a Complex Number · hard · theory
For a non-zero complex number $z$, let $\arg(z)$ denote the principal argument with $-\pi < \arg(z) \le \pi$. Then, which of the following statement(s) is (are) FALSE?
A. $\arg(-1-i) = \frac{\pi}{4}$, where $i = \sqrt{-1}$ ✓ Correct
B. The function $f : \mathbb{R} \to (-\pi, \pi]$, defined by $f(t) = \arg(-1+it)$ for all $t \in \mathbb{R}$, is continuous at all points of $\mathbb{R}$, where $i = \sqrt{-1}$ ✓ Correct
C. For any two non-zero complex numbers $z_1$ and $z_2$, $\arg\left(\frac{z_1}{z_2}\right) - \arg(z_1) + \arg(z_2)$ is an integer multiple of $2\pi$
D. For any three given distinct complex numbers $z_1$, $z_2$ and $z_3$, the locus of the point $z$ satisfying the condition $\arg\left(\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\right) = \pi$ lies on a straight line ✓ Correct
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.
Q8 — Argument of a Complex Number · medium · theory
Let $z_1$ and $z_2$ be two distinct complex numbers and let $z = (1-t)z_1 + tz_2$ for some real number $t$ with $0 < t < 1$. If $\arg(w)$ denotes the principal argument of a non-zero complex number $w$, then
A. $|z - z_1| + |z - z_2| = |z_1 - z_2|$ ✓ Correct
B. $\arg(z - z_1) = \arg(z - z_2)$
C. $\begin{vmatrix} z - z_1 & \bar{z} - \bar{z}_1 \\ z_2 - z_1 & \bar{z}_2 - \bar{z}_1 \end{vmatrix} = 0$ ✓ Correct
D. $\arg(z - z_1) = \arg(z_2 - z_1)$ ✓ Correct
Solution: Since $z = (1-t)z_1 + tz_2$ with $0 < t < 1$, the point $z$ lies on the segment joining $z_1$ and $z_2$, so $|z-z_1| + |z-z_2| = |z_1-z_2|$ — (a) is true. Also $z - z_1 = t(z_2 - z_1)$ with $t > 0$, so $\arg(z-z_1) = \arg(z_2-z_1)$ — (d) is true — and $\frac{z-z_1}{z_2-z_1} = t$ is purely real, which makes the determinant in (c) vanish. But $z - z_2 = (t-1)(z_2 - z_1)$ is a negative real multiple of $z_2 - z_1$, so $\arg(z-z_2)$ differs from $\arg(z-z_1)$ by $\pi$ — (b) is false.