Class 11 Maths Exercise 4.2: Argand Plane and Polar Form
Overview
This page provides comprehensive Class 11 Ch 4: Complex Numbers - Exercise 4.2 Practice. Free NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations Exercise 4-2. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.
Complex Numbers: Argand Plane and Polar Representation
About Class 11 Maths Complex Numbers And Quadratic Equations Exercise 4.2
This NCERT Class 11 Maths exercise focuses on Complex Numbers: Argand Plane and Polar Representation from the chapter Complex Numbers And Quadratic Equations. Revise the relevant definitions, standard results, notation, and worked examples before attempting the questions.
Attempt each question independently and write the reasoning in complete steps. Check symbols, substitutions, diagrams, and final conditions carefully, then use the step-by-step solution to identify exactly what needs revision.
Q1
00:00
Find the modulus and the argument of the complex number $z = -1 - i\sqrt{3}$.
Let $z = r(\cos \theta + i \sin \theta)$. Here $x = -1, y = -\sqrt{3}$.
Modulus $r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2$.
Let $\alpha$ be the acute angle such that $\tan \alpha = |\frac{y}{x}| = |\frac{-\sqrt{3}}{-1}| = \sqrt{3}$. So $\alpha = \frac{\pi}{3}$.
Since $x < 0$ and $y < 0$, the point lies in the third quadrant.
Argument $\theta = -(\pi - \alpha) = -(\pi - \frac{\pi}{3}) = -\frac{2\pi}{3}$.
Modulus = 2, Argument = $-\frac{2\pi}{3}$
Q2
00:00
Find the modulus and the argument of the complex number $z = -\sqrt{3} + i$.
Let $z = r(\cos \theta + i \sin \theta)$. Here $x = -\sqrt{3}, y = 1$.
Modulus $r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$.
Let $\alpha$ be acute angle: $\tan \alpha = |\frac{1}{-\sqrt{3}}| = \frac{1}{\sqrt{3}}$. So $\alpha = \frac{\pi}{6}$.
Since $x < 0, y > 0$, point is in second quadrant.
Argument $\theta = \pi - \alpha = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Modulus = 2, Argument = $\frac{5\pi}{6}$
Q3
00:00
Convert the complex number $1 - i$ to polar form.
Let $z = 1 - i$. $r = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
$\tan \alpha = |\frac{-1}{1}| = 1 \Rightarrow \alpha = \frac{\pi}{4}$.
Since $x > 0, y < 0$, point is in fourth quadrant. $\theta = -\alpha = -\frac{\pi}{4}$.
Polar form: $z = r(\cos \theta + i \sin \theta) = \sqrt{2}(\cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4}))$.
$= \sqrt{2}(\cos \frac{\pi}{4} - i \sin \frac{\pi}{4})$.
$\sqrt{2}(\cos \frac{\pi}{4} - i \sin \frac{\pi}{4})$
Q4
00:00
Convert the complex number $-1 + i$ to polar form.
Let $z = -1 + i$. $r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.
$\tan \alpha = |\frac{1}{-1}| = 1 \Rightarrow \alpha = \frac{\pi}{4}$.
Since $x < 0, y > 0$, point is in second quadrant. $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Polar form: $\sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})$.
$\sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})$
Q5
00:00
Convert the complex number $-1 - i$ to polar form.
Let $z = -1 - i$. $r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$.
$\tan \alpha = |\frac{-1}{-1}| = 1 \Rightarrow \alpha = \frac{\pi}{4}$.
Since $x < 0, y < 0$, point is in third quadrant. $\theta = -(\pi - \frac{\pi}{4}) = -\frac{3\pi}{4}$.
Polar form: $\sqrt{2}(\cos(-\frac{3\pi}{4}) + i \sin(-\frac{3\pi}{4})) = \sqrt{2}(\cos \frac{3\pi}{4} - i \sin \frac{3\pi}{4})$.
$\sqrt{2}(\cos \frac{3\pi}{4} - i \sin \frac{3\pi}{4})$
Q6
00:00
Convert the complex number $-3$ to polar form.
Let $z = -3 + 0i$. $r = \sqrt{(-3)^2} = 3$.
Point $(-3, 0)$ lies on the negative real axis.
Argument $\theta = \pi$.
Polar form: $3(\cos \pi + i \sin \pi)$.
$3(\cos \pi + i \sin \pi)$
Q7
00:00
Convert the complex number $\sqrt{3} + i$ to polar form.
Let $z = \sqrt{3} + i$. $r = \sqrt{3 + 1} = 2$.
$\tan \alpha = |\frac{1}{\sqrt{3}}| \Rightarrow \alpha = \frac{\pi}{6}$.
Since $x > 0, y > 0$, point is in first quadrant. $\theta = \frac{\pi}{6}$.
Polar form: $2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.
$2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$
Q8
00:00
Convert the complex number $i$ to polar form.
Let $z = 0 + i$. $r = 1$.
Point $(0, 1)$ lies on the positive imaginary axis.
Argument $\theta = \frac{\pi}{2}$.
Polar form: $1(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$.
$\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$