RBSE Solutions Class 11 Maths Chapter 5 Complex Numbers Exercise 5.2

Question 1. Find the arguments of the following numbers:
(i) \( \frac{1+i}{1-i} \)
(ii) \( -1+\sqrt{3}i \)
(iii) \( \frac{5+i\sqrt{3}}{4-i2\sqrt{3}} \)
Answer:
(i) Let \( z = \frac{1+i}{1-i} \). To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is \( (1+i) \).
\( z = \frac{1+i}{1-i} \times \frac{1+i}{1+i} \)
\( z = \frac{(1+i)^2}{1^2 - i^2} \)
\( z = \frac{1^2 + i^2 + 2i}{1 - (-1)} \)
\( z = \frac{1 - 1 + 2i}{1 + 1} \)
\( z = \frac{2i}{2} \)
\( z = i \)
We can write this as \( z = 0 + 1i \). Comparing with \( a+ib \), we have \( a=0 \) and \( b=1 \). Since \( a=0 \) and \( b=1 \), the complex number lies on the positive imaginary axis, which is in the first quadrant.
The argument \( \theta \) is given by \( \tan^{-1} \left( \frac{b}{a} \right) \).
\( \theta = \tan^{-1} \left( \frac{1}{0} \right) \)
\( \theta = \tan^{-1} (\infty) \)
\( \theta = \frac{\pi}{2} \)

(ii) Let \( z = -1+\sqrt{3}i \). Comparing with \( a+ib \), we have \( a=-1 \) and \( b=\sqrt{3} \).
Since \( a \) is negative and \( b \) is positive, the complex number \( z \) lies in the second quadrant.
In the second quadrant, the argument \( \theta \) is found using the formula \( \theta = \pi - \tan^{-1} \left| \frac{b}{a} \right| \).
\( \theta = \pi - \tan^{-1} \left| \frac{\sqrt{3}}{-1} \right| \)
\( \theta = \pi - \tan^{-1} (\sqrt{3}) \)
We know that \( \tan \frac{\pi}{3} = \sqrt{3} \), so \( \tan^{-1} (\sqrt{3}) = \frac{\pi}{3} \).
\( \theta = \pi - \frac{\pi}{3} \)
\( \theta = \frac{3\pi - \pi}{3} \)
\( \theta = \frac{2\pi}{3} \)

(iii) Let \( z = \frac{5+i\sqrt{3}}{4-i2\sqrt{3}} \). To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is \( (4+i2\sqrt{3}) \).
\( z = \frac{5+i\sqrt{3}}{4-i2\sqrt{3}} \times \frac{4+i2\sqrt{3}}{4+i2\sqrt{3}} \)
\( z = \frac{(5)(4) + (5)(i2\sqrt{3}) + (i\sqrt{3})(4) + (i\sqrt{3})(i2\sqrt{3})}{4^2 - (i2\sqrt{3})^2} \)
\( z = \frac{20 + i10\sqrt{3} + i4\sqrt{3} + i^2 (2)(\sqrt{3})(\sqrt{3})}{16 - i^2 (4)(3)} \)
\( z = \frac{20 + i14\sqrt{3} - 6}{16 - (-1)(12)} \)
\( z = \frac{14 + i14\sqrt{3}}{16 + 12} \)
\( z = \frac{14 + i14\sqrt{3}}{28} \)
Now, divide each term by 28:
\( z = \frac{14}{28} + i\frac{14\sqrt{3}}{28} \)
\( z = \frac{1}{2} + i\frac{\sqrt{3}}{2} \)
Comparing with \( a+ib \), we have \( a=\frac{1}{2} \) and \( b=\frac{\sqrt{3}}{2} \).
Since both \( a \) and \( b \) are positive, the complex number \( z \) lies in the first quadrant.
The argument \( \theta \) is given by \( \tan^{-1} \left( \frac{b}{a} \right) \).
\( \theta = \tan^{-1} \left( \frac{\sqrt{3}/2}{1/2} \right) \)
\( \theta = \tan^{-1} (\sqrt{3}) \)
We know that \( \tan \frac{\pi}{3} = \sqrt{3} \), so \( \tan^{-1} (\sqrt{3}) = \frac{\pi}{3} \).
\( \theta = \frac{\pi}{3} \)
In simple words: To find the argument, first simplify the complex number into the form \( a+ib \). Then, look at the signs of \( a \) and \( b \) to know which quadrant the number is in. Use the correct formula for that quadrant involving \( \tan^{-1} \left| \frac{b}{a} \right| \) to calculate the angle.

🎯 Exam Tip: Always simplify the complex number to the form \( a+ib \) before finding its argument. Pay close attention to the quadrant where the complex number lies to apply the correct argument formula.

Question 2. Express the following complex numbers into the polar form:
(i) \( \frac{1+i}{\sqrt{2}} \)
(ii) \( \frac{1+7i}{(2-i)^2} \)
(iii) \( \sin \frac{\pi}{3} + i \cos \frac{\pi}{3} \)
Answer:
(i) Let \( z = \frac{1+i}{\sqrt{2}} \). We can write this as \( z = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \).
Comparing with \( a+ib \), we have \( a=\frac{1}{\sqrt{2}} \) and \( b=\frac{1}{\sqrt{2}} \).
Since both \( a \) and \( b \) are positive, the complex number \( z \) lies in the first quadrant.
First, find the modulus \( r \):
\( r = \sqrt{a^2+b^2} = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} \)
\( r = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1 \)
Next, find the argument \( \theta \):
\( \theta = \tan^{-1} \left( \frac{b}{a} \right) = \tan^{-1} \left( \frac{1/\sqrt{2}}{1/\sqrt{2}} \right) \)
\( \theta = \tan^{-1} (1) \)
\( \theta = \frac{\pi}{4} \)
The polar form is \( z = r(\cos \theta + i \sin \theta) \).
\( z = 1\left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right) \)

(ii) Let \( z = \frac{1+7i}{(2-i)^2} \). First, simplify the denominator:
\( (2-i)^2 = 2^2 + i^2 - 2(2)(i) = 4 - 1 - 4i = 3 - 4i \).
So, \( z = \frac{1+7i}{3-4i} \). To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is \( (3+4i) \).
\( z = \frac{1+7i}{3-4i} \times \frac{3+4i}{3+4i} \)
\( z = \frac{(1)(3) + (1)(4i) + (7i)(3) + (7i)(4i)}{3^2 - (4i)^2} \)
\( z = \frac{3 + 4i + 21i + 28i^2}{9 - 16i^2} \)
\( z = \frac{3 + 25i - 28}{9 - 16(-1)} \)
\( z = \frac{-25 + 25i}{9 + 16} \)
\( z = \frac{-25 + 25i}{25} \)
Now, divide each term by 25:
\( z = \frac{-25}{25} + i\frac{25}{25} \)
\( z = -1 + i \)
Comparing with \( a+ib \), we have \( a=-1 \) and \( b=1 \).
Since \( a \) is negative and \( b \) is positive, the complex number \( z \) lies in the second quadrant.
First, find the modulus \( r \):
\( r = \sqrt{a^2+b^2} = \sqrt{(-1)^2 + (1)^2} \)
\( r = \sqrt{1+1} = \sqrt{2} \)
Next, find the argument \( \theta \). In the second quadrant, \( \theta = \pi - \tan^{-1} \left| \frac{b}{a} \right| \).
\( \theta = \pi - \tan^{-1} \left| \frac{1}{-1} \right| \)
\( \theta = \pi - \tan^{-1} (1) \)
\( \theta = \pi - \frac{\pi}{4} \)
\( \theta = \frac{3\pi}{4} \)
The polar form is \( z = r(\cos \theta + i \sin \theta) \).
\( z = \sqrt{2}\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right) \)

(iii) Let \( z = \sin \frac{\pi}{3} + i \cos \frac{\pi}{3} \).
The standard polar form is \( r(\cos \theta + i \sin \theta) \). We need to change the sine and cosine terms to match this order.
We use the trigonometric identities: \( \sin x = \cos \left( \frac{\pi}{2} - x \right) \) and \( \cos x = \sin \left( \frac{\pi}{2} - x \right) \).
Here, \( x = \frac{\pi}{3} \). So, the new angle \( \theta' \) will be \( \frac{\pi}{2} - \frac{\pi}{3} \).
\( \theta' = \frac{3\pi - 2\pi}{6} = \frac{\pi}{6} \)
Substitute this new angle back into the expression for \( z \):
\( z = \cos \left( \frac{\pi}{2} - \frac{\pi}{3} \right) + i \sin \left( \frac{\pi}{2} - \frac{\pi}{3} \right) \)
\( z = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \)
Comparing this with the standard polar form \( r(\cos \theta + i \sin \theta) \), we find that \( r=1 \) and \( \theta = \frac{\pi}{6} \).
The polar form is \( z = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \).
In simple words: To convert a complex number to polar form, first write it as \( a+ib \). Then, find its length (modulus \( r \)) and its angle (argument \( \theta \)). The polar form is then \( r \) multiplied by \( (\cos \theta + i \sin \theta) \). If the complex number is already in a sine-cosine format but not in the standard order, use angle conversion formulas to fix it.

🎯 Exam Tip: Remember that polar form is always \( r(\cos \theta + i \sin \theta) \). If sine and cosine are swapped, use \( \sin x = \cos (\frac{\pi}{2}-x) \) and \( \cos x = \sin (\frac{\pi}{2}-x) \) to adjust the angle and order.

Question 3. If \( z_1 \) and \( z_2 \) are two non-zero complex numbers, then prove that \( \arg z_1z_2 = \arg z_1 - \arg z_2 \).
Answer: Let the two non-zero complex numbers be represented in their polar forms as:
\( z_1 = r_1 (\cos \theta_1 + i \sin \theta_1) \)
And \( z_2 = r_2 (\cos \theta_2 + i \sin \theta_2) \)
From the definition of \( z_2 \), its conjugate, denoted as \( \overline{z_2} \), will be:
\( \overline{z_2} = r_2 (\cos \theta_2 - i \sin \theta_2) \)
Now, we multiply \( z_1 \) by the conjugate of \( z_2 \), which the question refers to as \( z_1z_2 \) in context of the proof:
\( z_1 \overline{z_2} = r_1 (\cos \theta_1 + i \sin \theta_1) \cdot r_2 (\cos \theta_2 - i \sin \theta_2) \)
We multiply the moduli and the complex parts:
\( z_1 \overline{z_2} = r_1 r_2 [ (\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 - i \sin \theta_2) ] \)
\( z_1 \overline{z_2} = r_1 r_2 [ \cos \theta_1 \cos \theta_2 - i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 - i^2 \sin \theta_1 \sin \theta_2 ] \)
Since \( i^2 = -1 \), we substitute this into the expression:
\( z_1 \overline{z_2} = r_1 r_2 [ \cos \theta_1 \cos \theta_2 - i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 ] \)
Group the real and imaginary parts:
\( z_1 \overline{z_2} = r_1 r_2 [ (\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2) + i (\sin \theta_1 \cos \theta_2 - \cos \theta_1 \sin \theta_2) ] \)
Using the trigonometric identities for \( \cos(A-B) \) and \( \sin(A-B) \):
\( \cos(\theta_1 - \theta_2) = \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 \)
\( \sin(\theta_1 - \theta_2) = \sin \theta_1 \cos \theta_2 - \cos \theta_1 \sin \theta_2 \)
Substitute these identities back into the expression for \( z_1 \overline{z_2} \):
\( z_1 \overline{z_2} = r_1 r_2 \{ \cos (\theta_1 - \theta_2) + i \sin (\theta_1 - \theta_2) \} \)
The argument of a complex number in polar form \( R(\cos \phi + i \sin \phi) \) is \( \phi \).
Therefore, the argument of \( z_1 \overline{z_2} \) is \( (\theta_1 - \theta_2) \).
Since \( \theta_1 = \arg z_1 \) and \( \theta_2 = \arg z_2 \), we can write:
\( \arg (z_1 \overline{z_2}) = \arg z_1 - \arg z_2 \)
Hence Proved.
In simple words: When you multiply one complex number by the conjugate of another complex number, the argument of the resulting number is found by subtracting the argument of the second number from the argument of the first number. This rule makes calculating the angle of the product simpler.

🎯 Exam Tip: When proving properties related to arguments of complex numbers, always start by expressing the complex numbers in their polar forms \( r(\cos \theta + i \sin \theta) \). This makes trigonometric identities easy to apply.