Complex Number in Iota Form — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ Complex Number in Iota Form MCQs with step-by-step solutions (10 questions). Part of Complex Numbers. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Complex Number in Iota Form · medium · numerical
Let $z \in \mathbb{C}$ with $\text{Im}(z) = 10$ and it satisfies $\frac{2z - n}{2z + n} = 2i - 1$ for some natural number $n$. Then:
A. $n = 20$ and $\text{Re}(z) = -10$
B. $n = 40$ and $\text{Re}(z) = 10$
C. $n = 40$ and $\text{Re}(z) = -10$ ✓ Correct
D. $n = 20$ and $\text{Re}(z) = 10$
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 $\text{Re}(z) = -10$.
Q2 — Complex Number in Iota Form · easy · numerical
All the points in the set $S = \left\{ \dfrac{\alpha + i}{\alpha - i} : \alpha \in \mathbb{R} \right\}$ $(i = \sqrt{-1})$ lie on a:
A. circle whose radius is $\sqrt{2}$
B. straight line whose slope is $-1$
C. circle whose radius is $1$ ✓ Correct
D. straight line whose slope is $1$
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.)
Q3 — Complex Number in Iota Form · medium · numerical
Let $z \in \mathbb{C}$ be such that $|z| < 1$. If $\omega = \dfrac{5 + 3z}{5(1 - z)}$, then:
A. $4\,\text{Im}(\omega) > 5$
B. $5\,\text{Re}(\omega) > 1$ ✓ Correct
C. $5\,\text{Im}(\omega) < 1$
D. $5\,\text{Re}(\omega) > 4$
Solution: From $5\omega(1 - z) = 5 + 3z$ we get $z(3 + 5\omega) = 5\omega - 5$, so $|3 + 5\omega|\,|z| = 5|\omega - 1|$. Since $|z| < 1$, $|3 + 5\omega| > 5|\omega - 1|$, i.e. $\left|\omega + \frac{3}{5}\right| > |\omega - 1|$. Writing $\omega = x + iy$: $\left(x + \frac{3}{5}\right)^2 + y^2 > (x-1)^2 + y^2 \Rightarrow \frac{16}{5}x > \frac{16}{25} \Rightarrow x > \frac{1}{5}$. Hence $5\,\text{Re}(\omega) > 1$.
Q4 — Complex Number in Iota Form · medium · numerical
Let $\left(-2 - \dfrac{1}{3}i\right)^3 = \dfrac{x + iy}{27}$ $(i = \sqrt{-1})$, where $x$ and $y$ are real numbers, then $y - x$ equals:
A. $91$ ✓ Correct
B. $85$
C. $-85$
D. $-91$
Solution: $x + iy = 27\left(-2 - \frac{i}{3}\right)^3 = 27 \cdot \left(-\frac{1}{27}\right)(6 + i)^3 = -(6+i)^3$. Now $(6+i)^3 = 216 + 3(36)i + 3(6)i^2 + i^3 = 216 + 108i - 18 - i = 198 + 107i$. So $x = -198$, $y = -107$, and $y - x = -107 + 198 = 91$.
Q5 — Complex Number in Iota Form · medium · numerical
Let $A = \left\{ \theta \in \left(-\dfrac{\pi}{2}, \pi\right) : \dfrac{3 + 2i\sin\theta}{1 - 2i\sin\theta} \text{ is purely imaginary} \right\}$. Then, the sum of the elements in $A$ is:
A. $\dfrac{3\pi}{4}$
B. $\dfrac{5\pi}{6}$
C. $\pi$
D. $\dfrac{2\pi}{3}$ ✓ Correct
Solution: Rationalising, $\frac{3 + 2i\sin\theta}{1 - 2i\sin\theta} \cdot \frac{1 + 2i\sin\theta}{1 + 2i\sin\theta} = \frac{3 - 4\sin^2\theta}{1 + 4\sin^2\theta} + \frac{8\sin\theta}{1 + 4\sin^2\theta}i$. Purely imaginary $\Rightarrow 3 - 4\sin^2\theta = 0 \Rightarrow \sin\theta = \pm\frac{\sqrt{3}}{2}$. In $\left(-\frac{\pi}{2}, \pi\right)$ this gives $\theta = -\frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3}$, whose sum is $\frac{2\pi}{3}$.
Q6 — Complex Number in Iota Form · easy · numerical
A value of $\theta$ for which $\dfrac{2 + 3i\sin\theta}{1 - 2i\sin\theta}$ is purely imaginary, is:
A. $\dfrac{\pi}{3}$
B. $\dfrac{\pi}{6}$
C. $\sin^{-1}\left(\dfrac{\sqrt{3}}{4}\right)$
D. $\sin^{-1}\left(\dfrac{1}{\sqrt{3}}\right)$ ✓ Correct
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.
Q7 — Complex Number in Iota Form · medium · numerical
If $\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix} = x + iy$, then:
A. $x = 3,\ y = 1$
B. $x = 1,\ y = 1$
C. $x = 0,\ y = 3$
D. $x = 0,\ y = 0$ ✓ Correct
Solution: Multiply column $C_2 = (-3i,\ 3i,\ 3)$ by $i$: $i\,C_2 = (3,\ -3,\ 3i) = 3(1,\ -1,\ i) = 3C_3$. So $C_2$ and $C_3$ are proportional, which makes the determinant zero. Hence $x + iy = 0$, giving $x = 0,\ y = 0$.
Q8 — Complex Number in Iota Form · easy · numerical
The value of the sum $\sum_{n=1}^{13} (i^n + i^{n+1})$, where $i = \sqrt{-1}$, equals:
A. $i$
B. $i - 1$ ✓ Correct
C. $-i$
D. $0$
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 $(1+i)i = i + i^2 = i - 1$.
Q9 — Complex Number in Iota Form · easy · numerical
The smallest positive integer $n$ for which $\left(\dfrac{1+i}{1-i}\right)^n = 1$, is:
A. $8$
B. $16$
C. $12$
D. None of these ✓ Correct
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.
Q10 — Complex Number in Iota Form · hard · numerical
Let $a, b, x$ and $y$ be real numbers such that $a - b = 1$ and $y \neq 0$. If the complex number $z = x + iy$ satisfies $\text{Im}\left(\dfrac{az + b}{z + 1}\right) = y$, then which of the following is(are) possible value(s) of $x$?
A. $1 - \sqrt{1 + y^2}$
B. $-1 - \sqrt{1 - y^2}$ ✓ Correct
C. $1 + \sqrt{1 + y^2}$
D. $-1 + \sqrt{1 - y^2}$ ✓ Correct
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 $x = -1 \pm \sqrt{1 - y^2}$.