GSEB Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations G Exercise 5.2

Find the modulus and the argument of each of the complex numbers in questions 1 to 2:

Question 1. \( z = -1 - i\sqrt{3} \)
Answer: Let \( z = -1 - i\sqrt{3} = r(\cos \theta + i \sin \theta) \).
Here, we compare the real and imaginary parts:
\( r \cos \theta = -1 \)
\( r \sin \theta = -\sqrt{3} \)
To find the modulus \( r \), we square both equations and add them:
\( (r \cos \theta)^2 + (r \sin \theta)^2 = (-1)^2 + (-\sqrt{3})^2 \)
\( r^2 \cos^2 \theta + r^2 \sin^2 \theta = 1 + 3 \)
\( r^2 (\cos^2 \theta + \sin^2 \theta) = 4 \)
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( r^2 = 4 \)
\( r = \sqrt{4} = 2 \) (as \( r \) must be a non-negative value).
Now, to find the argument \( \theta \), we use the values of \( r \cos \theta \) and \( r \sin \theta \):
\( \cos \theta = \frac{-1}{r} = \frac{-1}{2} \)
\( \sin \theta = \frac{-\sqrt{3}}{r} = \frac{-\sqrt{3}}{2} \)
Since both \( \cos \theta \) and \( \sin \theta \) are negative, the complex number \( z \) lies in the third quadrant.
The reference angle \( \alpha \) for \( \tan \alpha = \left| \frac{\sin \theta}{\cos \theta} \right| = \left| \frac{-\sqrt{3}/2}{-1/2} \right| = \sqrt{3} \) is \( \alpha = 60^\circ \) or \( \frac{\pi}{3} \) radians.
In the third quadrant, the argument \( \theta \) is given by \( \theta = -180^\circ + \alpha \) or \( \theta = -\pi + \alpha \).
So, \( \theta = -180^\circ + 60^\circ = -120^\circ \).
In radians, \( \theta = -\pi + \frac{\pi}{3} = -\frac{3\pi}{3} + \frac{\pi}{3} = -\frac{2\pi}{3} \).
Thus, the modulus \( |z| = 2 \) and the argument \( \arg z = -\frac{2\pi}{3} \).
In simple words: To find the modulus, square the real and imaginary parts, add them, and take the square root. To find the argument, look at the signs of the real and imaginary parts to determine the quadrant, then use tangent to find the angle.

Exam Tip: Always remember that the modulus \(r\) must be a positive value. When determining the argument \( \theta \), carefully check the signs of both \( \cos \theta \) and \( \sin \theta \) to correctly identify the quadrant where \( \theta \) lies, then use the appropriate formula to calculate the principal argument (usually in the range \( (-\pi, \pi] \)).

Question 2. \( z = - \sqrt{3} + i \)
Answer: Let \( z = -\sqrt{3} + i = r(\cos \theta + i \sin \theta) \).
Here, we compare the real and imaginary parts:
\( r \cos \theta = -\sqrt{3} \)
\( r \sin \theta = 1 \)
To find the modulus \( r \), we square both equations and add them:
\( (r \cos \theta)^2 + (r \sin \theta)^2 = (-\sqrt{3})^2 + (1)^2 \)
\( r^2 \cos^2 \theta + r^2 \sin^2 \theta = 3 + 1 \)
\( r^2 (\cos^2 \theta + \sin^2 \theta) = 4 \)
\( r^2 = 4 \)
\( r = \sqrt{4} = 2 \).
Now, to find the argument \( \theta \):
\( \cos \theta = \frac{-\sqrt{3}}{r} = \frac{-\sqrt{3}}{2} \)
\( \sin \theta = \frac{1}{r} = \frac{1}{2} \)
Since \( \cos \theta \) is negative and \( \sin \theta \) is positive, the complex number \( z \) lies in the second quadrant.
The reference angle \( \alpha \) for \( \tan \alpha = \left| \frac{\sin \theta}{\cos \theta} \right| = \left| \frac{1/2}{-\sqrt{3}/2} \right| = \frac{1}{\sqrt{3}} \) is \( \alpha = 30^\circ \) or \( \frac{\pi}{6} \) radians.
In the second quadrant, the argument \( \theta \) is given by \( \theta = 180^\circ - \alpha \) or \( \theta = \pi - \alpha \).
So, \( \theta = 180^\circ - 30^\circ = 150^\circ \).
In radians, \( \theta = \pi - \frac{\pi}{6} = \frac{6\pi}{6} - \frac{\pi}{6} = \frac{5\pi}{6} \).
Thus, the modulus \( |z| = 2 \) and the argument \( \arg z = \frac{5\pi}{6} \).
In simple words: First, find the distance from the origin (modulus) by squaring the real and imaginary parts, adding them, and taking the square root. Next, determine the angle (argument) by seeing which quadrant the number falls into and using basic trigonometry.

Exam Tip: When \(r\) is calculated, it's always positive. Ensure you use the correct signs for the real and imaginary parts to accurately place the complex number in its quadrant, which is crucial for determining the principal argument \( \theta \).

Convert each of the complex numbers given in questions 3 to 8 in the polar form:

Question 3. \( 1 - i \)
Answer: Let \( z = 1 - i = r(\cos \theta + i \sin \theta) \).
Comparing the real and imaginary parts:
\( r \cos \theta = 1 \)
\( r \sin \theta = -1 \)
To find the modulus \( r \):
\( r^2 = (1)^2 + (-1)^2 = 1 + 1 = 2 \)
\( r = \sqrt{2} \).
To find the argument \( \theta \):
\( \cos \theta = \frac{1}{\sqrt{2}} \)
\( \sin \theta = \frac{-1}{\sqrt{2}} \)
Since \( \cos \theta \) is positive and \( \sin \theta \) is negative, \( z \) lies in the fourth quadrant.
The reference angle \( \alpha \) for \( \tan \alpha = \left| \frac{-1/ \sqrt{2}}{1 / \sqrt{2}} \right| = 1 \) is \( \alpha = 45^\circ \) or \( \frac{\pi}{4} \) radians.
In the fourth quadrant, \( \theta = - \alpha \).
So, \( \theta = -45^\circ \) or \( -\frac{\pi}{4} \) radians.
The polar form of \( 1 - i \) is \( \sqrt{2} \left[ \cos \left( -\frac{\pi}{4} \right) + i \sin \left( -\frac{\pi}{4} \right) \right] \).
In simple words: Convert the complex number into its polar form. First, calculate \(r\) (the modulus) using the Pythagorean theorem. Then, determine \( \theta \) (the argument) by considering the quadrant and finding the angle from the positive x-axis.

Exam Tip: When converting to polar form, remember to always use the principal argument, which lies between \( -\pi \) and \( \pi \) (or \( -180^\circ \) and \( 180^\circ \)). Pay close attention to the quadrant to get the correct sign for the argument.

Question 4. \( -1 + i \)
Answer: Let \( z = -1 + i = r(\cos \theta + i \sin \theta) \).
Comparing the real and imaginary parts:
\( r \cos \theta = -1 \)
\( r \sin \theta = 1 \)
To find the modulus \( r \):
\( r^2 = (-1)^2 + (1)^2 = 1 + 1 = 2 \)
\( r = \sqrt{2} \).
To find the argument \( \theta \):
\( \cos \theta = \frac{-1}{\sqrt{2}} \)
\( \sin \theta = \frac{1}{\sqrt{2}} \)
Since \( \cos \theta \) is negative and \( \sin \theta \) is positive, \( z \) lies in the second quadrant.
The reference angle \( \alpha \) for \( \tan \alpha = \left| \frac{1/\sqrt{2}}{-1/\sqrt{2}} \right| = 1 \) is \( \alpha = 45^\circ \) or \( \frac{\pi}{4} \) radians.
In the second quadrant, \( \theta = 180^\circ - \alpha \) or \( \theta = \pi - \alpha \).
So, \( \theta = 180^\circ - 45^\circ = 135^\circ \).
In radians, \( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).
The polar form of \( -1 + i \) is \( \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) \).
In simple words: When a complex number is negative in the real part and positive in the imaginary part, it sits in the second quarter of the graph. Find its length from the center, then calculate its angle from the positive horizontal axis.

Exam Tip: A common mistake is to simply use \( \arctan(\frac{y}{x}) \) without considering the quadrant. Always visualize the complex number on the Argand plane to correctly determine the argument based on the signs of its real and imaginary parts.

Question 5. \( -3 \)
Answer: Let \( z = -3 \). This can be written as \( z = -3 + 0i \).
Let \( z = -3 + 0i = r(\cos \theta + i \sin \theta) \).
Comparing the real and imaginary parts:
\( r \cos \theta = -3 \)
\( r \sin \theta = 0 \)
To find the modulus \( r \):
\( r^2 = (-3)^2 + (0)^2 = 9 + 0 = 9 \)
\( r = \sqrt{9} = 3 \).
To find the argument \( \theta \):
\( \cos \theta = \frac{-3}{3} = -1 \)
\( \sin \theta = \frac{0}{3} = 0 \)
The angle \( \theta \) for which \( \cos \theta = -1 \) and \( \sin \theta = 0 \) is \( \theta = \pi \) radians (or \( 180^\circ \)).
The polar form of \( -3 \) is \( 3(\cos \pi + i \sin \pi) \).
In simple words: For a purely real negative number, the length from zero is its absolute value. Since it's on the negative real axis, its angle is 180 degrees.

Exam Tip: For purely real or purely imaginary complex numbers, you can often determine the modulus and argument by inspection. For example, a negative real number like -3 lies on the negative x-axis, so its argument is \( \pi \).

Question 6. \( -3 \)
Answer: Let \( z = -3 \). This can be written as \( z = -3 + 0i \).
Let \( z = -3 + 0i = r(\cos \theta + i \sin \theta) \).
Comparing the real and imaginary parts:
\( r \cos \theta = -3 \)
\( r \sin \theta = 0 \)
To find the modulus \( r \):
\( r^2 = (-3)^2 + (0)^2 = 9 + 0 = 9 \)
\( r = \sqrt{9} = 3 \).
To find the argument \( \theta \):
\( \cos \theta = \frac{-3}{3} = -1 \)
\( \sin \theta = \frac{0}{3} = 0 \)
The angle \( \theta \) for which \( \cos \theta = -1 \) and \( \sin \theta = 0 \) is \( \theta = \pi \) radians (or \( 180^\circ \)).
The polar form of \( -3 \) is \( 3(\cos \pi + i \sin \pi) \).
In simple words: When you have a negative number like -3, its length from the center is 3. Since it's directly left on the number line, its angle is \( \pi \) radians.

Exam Tip: Numbers on the real axis have arguments of \( 0 \) (for positive numbers) or \( \pi \) (for negative numbers). Numbers on the imaginary axis have arguments of \( \frac{\pi}{2} \) (for positive imaginary numbers) or \( -\frac{\pi}{2} \) (for negative imaginary numbers).

Question 7. \( \sqrt{3} + i \)
Answer: Let \( z = \sqrt{3} + i = r(\cos \theta + i \sin \theta) \).
Comparing the real and imaginary parts:
\( r \cos \theta = \sqrt{3} \)
\( r \sin \theta = 1 \)
To find the modulus \( r \):
\( r^2 = (\sqrt{3})^2 + (1)^2 = 3 + 1 = 4 \)
\( r = \sqrt{4} = 2 \).
To find the argument \( \theta \):
\( \cos \theta = \frac{\sqrt{3}}{2} \)
\( \sin \theta = \frac{1}{2} \)
Since both \( \cos \theta \) and \( \sin \theta \) are positive, \( z \) lies in the first quadrant.
The angle \( \theta \) for which \( \cos \theta = \frac{\sqrt{3}}{2} \) and \( \sin \theta = \frac{1}{2} \) is \( \theta = 30^\circ \) or \( \frac{\pi}{6} \) radians.
The polar form of \( \sqrt{3} + i \) is \( 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \).
In simple words: When both the real and imaginary parts are positive, the complex number is in the first quadrant. Calculate its length \(r\), then find its angle \( \theta \) directly using trigonometric values.

Exam Tip: Remember special angle values for sine and cosine (like 30°, 45°, 60°) as they frequently appear in complex number problems. This can help you quickly find the argument for numbers in the first quadrant.

Question 8. \( i \)
Answer: Let \( z = i \). This can be written as \( z = 0 + 1i \).
Let \( z = 0 + 1i = r(\cos \theta + i \sin \theta) \).
Comparing the real and imaginary parts:
\( r \cos \theta = 0 \)
\( r \sin \theta = 1 \)
To find the modulus \( r \):
\( r^2 = (0)^2 + (1)^2 = 0 + 1 = 1 \)
\( r = \sqrt{1} = 1 \).
To find the argument \( \theta \):
\( \cos \theta = \frac{0}{1} = 0 \)
\( \sin \theta = \frac{1}{1} = 1 \)
The angle \( \theta \) for which \( \cos \theta = 0 \) and \( \sin \theta = 1 \) is \( \theta = \frac{\pi}{2} \) radians (or \( 90^\circ \)).
The polar form of \( i \) is \( 1 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) \), which simplifies to \( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \).
In simple words: The imaginary unit \(i\) has a length of 1 from the origin. Since it lies on the positive imaginary axis, its angle is 90 degrees, or \( \frac{\pi}{2} \) radians.

Exam Tip: Purely imaginary numbers like \(i\) or \(-i\) are common. For \(i\), it's at \( (0, 1) \) on the Argand plane, so its modulus is 1 and its argument is \( \frac{\pi}{2} \). For \(-i\), it's at \( (0, -1) \), so its modulus is 1 and its argument is \( -\frac{\pi}{2} \).