RBSE Solutions Class 11 Maths Chapter 5 Complex Numbers Exercise 5.3

Question 1. Find the square root of the following complex numbers:
(i) -5 + 12i
(ii) 8 - 6i
(iii) -i
Answer: The formula to find the square root of a complex number \( a+ib \) is:
\( \sqrt{a+ib} = \pm \left[ \sqrt{\frac{\sqrt{a^2+b^2}+a}{2}} + i \sqrt{\frac{\sqrt{a^2+b^2}-a}{2}} \right] \)

(i) For \( -5 + 12i \): Here \( a = -5 \) and \( b = 12 \).
\( \sqrt{-5+12i} = \pm \left[ \sqrt{\frac{\sqrt{(-5)^2+12^2}+(-5)}{2}} + i \sqrt{\frac{\sqrt{(-5)^2+12^2}-(-5)}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{\sqrt{25+144}-5}{2}} + i \sqrt{\frac{\sqrt{25+144}+5}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{\sqrt{169}-5}{2}} + i \sqrt{\frac{\sqrt{169}+5}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{13-5}{2}} + i \sqrt{\frac{13+5}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{8}{2}} + i \sqrt{\frac{18}{2}} \right] \)
\( = \pm \left[ \sqrt{4} + i \sqrt{9} \right] \)
\( = \pm (2+3i) \)

(ii) For \( 8 - 6i \): Here \( a = 8 \) and \( b = -6 \).
\( \sqrt{8-6i} = \pm \left[ \sqrt{\frac{\sqrt{8^2+(-6)^2}+8}{2}} + i \sqrt{\frac{\sqrt{8^2+(-6)^2}-8}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{\sqrt{64+36}+8}{2}} + i \sqrt{\frac{\sqrt{64+36}-8}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{\sqrt{100}+8}{2}} + i \sqrt{\frac{\sqrt{100}-8}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{10+8}{2}} + i \sqrt{\frac{10-8}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{18}{2}} + i \sqrt{\frac{2}{2}} \right] \)
\( = \pm \left[ \sqrt{9} + i \sqrt{1} \right] \)
\( = \pm (3-i) \)

(iii) For \( -i \): This can be written as \( 0 - i \). Here \( a = 0 \) and \( b = -1 \).
\( \sqrt{-i} = \pm \left[ \sqrt{\frac{\sqrt{0^2+(-1)^2}+0}{2}} + i \sqrt{\frac{\sqrt{0^2+(-1)^2}-0}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{\sqrt{1}+0}{2}} + i \sqrt{\frac{\sqrt{1}-0}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{1}{2}} + i \sqrt{\frac{1}{2}} \right] \)
\( = \pm \frac{1}{\sqrt{2}} (1+i) \)
In simple words: To find the square root of a complex number, we use a special formula. We put the 'a' and 'b' values from the complex number into this formula and do the calculations step by step. This gives us two possible square roots, one positive and one negative.

🎯 Exam Tip: Remember the square root of a complex number always results in two values, positive and negative. Be careful with the signs of 'a' and 'b' when applying the formula.

Question 2. Find the value of \( \sqrt{4+3\sqrt{-20}} + \sqrt{4-3\sqrt{-20}} \)
Answer: We need to simplify \( \sqrt{-20} \) first.
\( \sqrt{-20} = \sqrt{20 \times (-1)} = \sqrt{20} \times \sqrt{-1} = \sqrt{4 \times 5} \times i = 2\sqrt{5}i \)
So the expression becomes: \( \sqrt{4+3(2\sqrt{5}i)} + \sqrt{4-3(2\sqrt{5}i)} = \sqrt{4+6\sqrt{5}i} + \sqrt{4-6\sqrt{5}i} \)
Now, let's find the square root of \( 4+6\sqrt{5}i \). Here \( a=4 \) and \( b=6\sqrt{5} \).
First, calculate \( \sqrt{a^2+b^2} \):
\( \sqrt{4^2+(6\sqrt{5})^2} = \sqrt{16+36 \times 5} = \sqrt{16+180} = \sqrt{196} = 14 \)
Using the formula for \( \sqrt{a+ib} \):
\( \sqrt{4+6\sqrt{5}i} = \pm \left[ \sqrt{\frac{14+4}{2}} + i \sqrt{\frac{14-4}{2}} \right] \)
\( = \pm \left[ \sqrt{\frac{18}{2}} + i \sqrt{\frac{10}{2}} \right] \)
\( = \pm \left[ \sqrt{9} + i \sqrt{5} \right] \)
\( = \pm (3+i\sqrt{5}) \) ...(i)

Similarly, for \( \sqrt{4-6\sqrt{5}i} \): Here \( a=4 \) and \( b=-6\sqrt{5} \). The value of \( \sqrt{a^2+b^2} \) is still 14.
Using the formula:
\( \sqrt{4-6\sqrt{5}i} = \pm \left[ \sqrt{\frac{14+4}{2}} - i \sqrt{\frac{14-4}{2}} \right] \)
\( = \pm \left[ \sqrt{9} - i \sqrt{5} \right] \)
\( = \pm (3-i\sqrt{5}) \) ...(ii)

Now we add the two square roots:
\( \sqrt{4+6\sqrt{5}i} + \sqrt{4-6\sqrt{5}i} \)
This can be \( (3+i\sqrt{5}) + (3-i\sqrt{5}) = 3+i\sqrt{5}+3-i\sqrt{5} = 6 \)
Or \( (3+i\sqrt{5}) + (-(3-i\sqrt{5})) = 3+i\sqrt{5}-3+i\sqrt{5} = 2i\sqrt{5} \)
Or \( -(3+i\sqrt{5}) + (3-i\sqrt{5}) = -3-i\sqrt{5}+3-i\sqrt{5} = -2i\sqrt{5} \)
Or \( -(3+i\sqrt{5}) + (-(3-i\sqrt{5})) = -3-i\sqrt{5}-3+i\sqrt{5} = -6 \)
So, the possible values are \( \pm 6, \pm 2i\sqrt{5} \).
In simple words: First, we change the square root of a negative number into a complex number using 'i'. Then, we find the square root of each complex part using a formula. Finally, we add these square roots together, which gives us four possible outcomes.

🎯 Exam Tip: When simplifying square roots of negative numbers, always convert \( \sqrt{-x} \) to \( i\sqrt{x} \). Remember that \( \pm \) means there are two possible values for each square root, leading to multiple outcomes when combining them.

Question 4. Prove that:
(i) \( 1 + \omega^n + \omega^{2n} = 0 \), whereas \( n = 2, 4 \)
(ii) \( 1 + \omega^n + \omega^{2n} = 3 \), whereas \( n \) is a multiple of 3.
Answer: We know that \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).

(i) Prove \( 1 + \omega^n + \omega^{2n} = 0 \), for \( n=2, 4 \)
When \( n=2 \):
\( 1 + \omega^2 + \omega^{2 \times 2} = 1 + \omega^2 + \omega^4 \)
\( = 1 + \omega^2 + \omega^3 \cdot \omega \)
\( = 1 + \omega^2 + 1 \cdot \omega \) (Since \( \omega^3 = 1 \))
\( = 1 + \omega + \omega^2 \)
\( = 0 \) (Since \( 1 + \omega + \omega^2 = 0 \))

When \( n=4 \):
\( 1 + \omega^4 + \omega^{2 \times 4} = 1 + \omega^4 + \omega^8 \)
\( = 1 + \omega^3 \cdot \omega + \omega^6 \cdot \omega^2 \)
\( = 1 + 1 \cdot \omega + (\omega^3)^2 \cdot \omega^2 \) (Since \( \omega^3 = 1 \))
\( = 1 + \omega + 1^2 \cdot \omega^2 \)
\( = 1 + \omega + \omega^2 \)
\( = 0 \) (Since \( 1 + \omega + \omega^2 = 0 \))
Hence, Proved.

(ii) Prove \( 1 + \omega^n + \omega^{2n} = 3 \), whereas \( n \) is a multiple of 3.
Let \( n = 3k \), where \( k \) is an integer.
\( 1 + \omega^{3k} + \omega^{2 \times 3k} = 1 + (\omega^3)^k + (\omega^3)^{2k} \)
\( = 1 + (1)^k + (1)^{2k} \) (Since \( \omega^3 = 1 \))
\( = 1 + 1 + 1 \)
\( = 3 \)
Hence, Proved.
In simple words: This problem uses the special properties of cube roots of unity, where \( \omega^3 \) equals 1 and \( 1+\omega+\omega^2 \) equals 0. We substitute the given values of 'n' into the expression and simplify it using these properties to show the required results.

🎯 Exam Tip: Always remember the fundamental properties of cube roots of unity: \( \omega^3 = 1 \) and \( 1+\omega+\omega^2 = 0 \). These are key to simplifying expressions involving \( \omega \).

Question 5. Prove that:
(i) If \( \omega \) is a cube root of unity, then \( \left( \frac{-1+i\sqrt{3}}{2} \right)^{29} + \left( \frac{-1-i\sqrt{3}}{2} \right)^{29} = -1 \).
(ii) \( (1 + 5\omega^2 + \omega)(1 + 5\omega + \omega^2)(5 + \omega + \omega^2) = 64 \).
Answer: We know that if \( \omega \) is a cube root of unity, then \( \omega = \frac{-1+i\sqrt{3}}{2} \) and \( \omega^2 = \frac{-1-i\sqrt{3}}{2} \). Also, \( \omega^3 = 1 \) and \( 1+\omega+\omega^2 = 0 \).

(i) Let's take the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = \left( \frac{-1+i\sqrt{3}}{2} \right)^{29} + \left( \frac{-1-i\sqrt{3}}{2} \right)^{29} \)
Substitute \( \omega \) and \( \omega^2 \):
\( = (\omega)^{29} + (\omega^2)^{29} \)
\( = \omega^{29} + \omega^{58} \)
Since \( \omega^3 = 1 \), we can simplify the powers:
\( \omega^{29} = \omega^{3 \times 9 + 2} = (\omega^3)^9 \cdot \omega^2 = 1^9 \cdot \omega^2 = \omega^2 \)
\( \omega^{58} = \omega^{3 \times 19 + 1} = (\omega^3)^{19} \cdot \omega^1 = 1^{19} \cdot \omega = \omega \)
So,
\( = \omega^2 + \omega \)
From the property \( 1+\omega+\omega^2 = 0 \), we know that \( \omega^2 + \omega = -1 \).
\( = -1 \)
\( = \text{R.H.S.} \)
Hence, Proved.

(ii) Let's take the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = (1 + 5\omega^2 + \omega)(1 + 5\omega + \omega^2)(5 + \omega + \omega^2) \)
Rearrange terms to use \( 1+\omega+\omega^2 = 0 \):
First factor: \( (1 + \omega + 5\omega^2) = (1 + \omega + \omega^2 + 4\omega^2) = (0 + 4\omega^2) = 4\omega^2 \)
Second factor: \( (1 + \omega^2 + 5\omega) = (1 + \omega + \omega^2 + 4\omega) = (0 + 4\omega) = 4\omega \)
Third factor: \( (5 + \omega + \omega^2) = (4 + 1 + \omega + \omega^2) = (4 + 0) = 4 \)
Now, multiply these simplified factors:
\( = (4\omega^2)(4\omega)(4) \)
\( = 64\omega^3 \)
Since \( \omega^3 = 1 \):
\( = 64 \times 1 \)
\( = 64 \)
\( = \text{R.H.S.} \)
Hence, Proved.
In simple words: We use the fundamental properties of the cube roots of unity, \( \omega^3=1 \) and \( 1+\omega+\omega^2=0 \). For part (i), we simplify the powers of \( \omega \) to show it equals -1. For part (ii), we rearrange the terms in each bracket to use the sum property, simplify them, and then multiply the results to get 64.

🎯 Exam Tip: When dealing with complex expressions involving \( \omega \), always simplify powers of \( \omega \) using \( \omega^3 = 1 \) and group terms to use the identity \( 1+\omega+\omega^2=0 \). This makes calculations much easier and faster.