Samacheer Kalvi Class 12 Maths Solutions Chapter 2 Complex Numbers Exercise 2.4

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.4

Question 1. Write the following in the rectangular form:
(i) \( \overline { (5+9i)+(2-4i) } \)
Answer:
(i) To write \( \overline { (5+9i)+(2-4i) } \) in rectangular form, we first add the complex numbers inside the bar.
\( (5+9i)+(2-4i) = (5+2) + (9-4)i = 7+5i \)
Now, we take the conjugate of the result:
\( \overline{7+5i} = 7-5i \)
The rectangular form is \( 7-5i \). Complex conjugates are useful for simplifying expressions involving imaginary numbers.
In simple words: First, add the two complex numbers. Then, change the sign of the imaginary part to find the conjugate, which gives the rectangular form.

🎯 Exam Tip: Remember that \( \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2} \), which can sometimes simplify calculations for sums.

(ii) \( \frac {10-5i}{6+2i} \)
Answer:
(ii) To write \( \frac {10-5i}{6+2i} \) in rectangular form, we multiply the numerator and denominator by the conjugate of the denominator.
The conjugate of \( 6+2i \) is \( 6-2i \).
\( \frac {10-5i}{6+2i} = \frac {10-5i}{6+2i} \times \frac {6-2i}{6-2i} \)
\( = \frac {(10-5i)(6-2i)}{(6+2i)(6-2i)} \)
We expand the numerator and denominator:
Numerator: \( (10-5i)(6-2i) = 60 - 20i - 30i + 10i^2 = 60 - 50i + 10(-1) = 60 - 50i - 10 = 50 - 50i \)
Denominator: \( (6+2i)(6-2i) = 6^2 - (2i)^2 = 36 - 4i^2 = 36 - 4(-1) = 36 + 4 = 40 \)
So, \( \frac {50-50i}{40} = \frac {50(1-i)}{40} = \frac {5(1-i)}{4} \)
This can be written as \( \frac{5}{4} - \frac{5}{4}i \). This process is similar to rationalizing denominators with square roots.
In simple words: Multiply the top and bottom of the fraction by the conjugate of the bottom part. Then, simplify the expression to get it into the form a + bi.

🎯 Exam Tip: When dividing complex numbers, always multiply by the conjugate of the denominator to make the denominator a real number.

(iii) \( \overline {3i} + \frac{2}{2-i} \)
Answer:
(iii) To write \( \overline {3i} + \frac{2}{2-i} \) in rectangular form, we handle each part separately.
First, find the conjugate of \( 3i \):
\( \overline{3i} = -3i \)
Next, simplify the fraction \( \frac{2}{2-i} \) by multiplying the numerator and denominator by the conjugate of \( 2-i \), which is \( 2+i \).
\( \frac{2}{2-i} = \frac{2}{2-i} \times \frac{2+i}{2+i} = \frac{2(2+i)}{(2-i)(2+i)} \)
\( = \frac{4+2i}{2^2 - i^2} = \frac{4+2i}{4 - (-1)} = \frac{4+2i}{5} = \frac{4}{5} + \frac{2}{5}i \)
Now, add the two simplified parts:
\( -3i + \left( \frac{4}{5} + \frac{2}{5}i \right) = \frac{4}{5} + \left( \frac{2}{5} - 3 \right)i \)
To combine the imaginary parts: \( \frac{2}{5} - 3 = \frac{2}{5} - \frac{15}{5} = \frac{2-15}{5} = \frac{-13}{5} \)
So, the expression in rectangular form is \( \frac{4}{5} - \frac{13}{5}i \). Combining the real and imaginary parts is key to getting the final form.
In simple words: Take the conjugate of the first part, then simplify the second part by multiplying by the conjugate of its denominator. Finally, add the real parts together and the imaginary parts together.

🎯 Exam Tip: When an expression has multiple operations, simplify each part step-by-step before combining them. Always ensure the final answer is in the \( a+bi \) form.

Question 2. If \( z = x + iy \), find the following in rectangular form.
(i) \( Re\left(\frac {1}{z}\right) \)
Answer:
(i) Given \( z = x + iy \). To find \( Re\left(\frac {1}{z}\right) \), we first find \( \frac {1}{z} \) in rectangular form.
\( \frac {1}{z} = \frac {1}{x+iy} \)
Multiply the numerator and denominator by the conjugate of \( x+iy \), which is \( x-iy \).
\( \frac {1}{x+iy} \times \frac {x-iy}{x-iy} = \frac {x-iy}{x^2 - (iy)^2} = \frac {x-iy}{x^2 - i^2y^2} = \frac {x-iy}{x^2 - (-1)y^2} = \frac {x-iy}{x^2+y^2} \)
This can be written as \( \frac {x}{x^2+y^2} - i\frac {y}{x^2+y^2} \).
The real part of \( \frac {1}{z} \) is the term without \( i \).
Therefore, \( Re\left(\frac {1}{z}\right) = \frac {x}{x^2+y^2} \). This calculation demonstrates how to isolate the real component after division.
In simple words: First, find the reciprocal of z. To do this, multiply the top and bottom by the conjugate of z. The part of the answer that does not have 'i' is the real part.

🎯 Exam Tip: Always remember that \( z \overline{z} = x^2+y^2 \) for \( z = x+iy \), which helps simplify the denominator quickly when finding reciprocals.

(ii) \( Re(i\overline{z}) \)
Answer:
(ii) Given \( z = x + iy \). We need to find \( Re(i\overline{z}) \).
First, find the conjugate of \( z \): \( \overline{z} = \overline{x+iy} = x-iy \).
Next, multiply \( i \) by \( \overline{z} \):
\( i\overline{z} = i(x-iy) = ix - i^2y = ix - (-1)y = ix + y \)
Rearrange this into rectangular form \( a+bi \): \( y + ix \).
The real part of \( y + ix \) is the term without \( i \).
Therefore, \( Re(i\overline{z}) = y \). This shows how the real part of an expression can change depending on the operations involved.
In simple words: First, find the conjugate of z. Then, multiply it by 'i'. The part of the result without 'i' is the real part.

🎯 Exam Tip: Be careful with the multiplication by \( i \), remembering that \( i^2 = -1 \). Always arrange the result into \( a+bi \) form before identifying the real or imaginary parts.

(iii) \( Im(3z + 4\overline{z} – 4i) \)
Answer:
(iii) Given \( z = x + iy \). We need to find \( Im(3z + 4\overline{z} – 4i) \).
First, substitute \( z = x+iy \) and \( \overline{z} = x-iy \) into the expression:
\( 3z = 3(x+iy) = 3x+3iy \)
\( 4\overline{z} = 4(x-iy) = 4x-4iy \)
Now, combine these with \( -4i \):
\( 3z + 4\overline{z} – 4i = (3x+3iy) + (4x-4iy) - 4i \)
Group the real and imaginary parts:
Real parts: \( 3x + 4x = 7x \)
Imaginary parts: \( 3iy - 4iy - 4i = (3y - 4y - 4)i = (-y - 4)i \)
So, the expression in rectangular form is \( 7x + (-y-4)i \).
The imaginary part of this expression is the coefficient of \( i \).
Therefore, \( Im(3z + 4\overline{z} – 4i) = -y-4 \), which can also be written as \( -(y+4) \). This calculation shows how properties of complex numbers help simplify larger expressions.
In simple words: Replace z with (x + iy) and its conjugate with (x - iy). Expand and group all the real parts together and all the imaginary parts together. The number multiplying 'i' in the final simplified expression is the imaginary part.

🎯 Exam Tip: When combining multiple complex numbers, always expand all terms first and then gather all real terms and all imaginary terms separately before identifying the final real or imaginary component.

Question 3. If \( z_1 = 2 − i \) and \( z_2 = -4 + 3i \), find the inverse of \( z_1 \), \( z_2 \) and \( \frac {z_1}{z_2} \)
Answer:
Given \( z_1 = 2 - i \) and \( z_2 = -4 + 3i \).
**1. Find the inverse of \( z_1 \):**
The inverse of \( z_1 \) is \( \frac{1}{z_1} \).
\( \frac{1}{z_1} = \frac{1}{2-i} \)
Multiply by the conjugate \( (2+i) \):
\( \frac{1}{2-i} \times \frac{2+i}{2+i} = \frac{2+i}{2^2 - i^2} = \frac{2+i}{4 - (-1)} = \frac{2+i}{5} = \frac{2}{5} + \frac{1}{5}i \)

**2. Find the inverse of \( z_2 \):**
The inverse of \( z_2 \) is \( \frac{1}{z_2} \).
\( \frac{1}{z_2} = \frac{1}{-4+3i} \)
Multiply by the conjugate \( (-4-3i) \):
\( \frac{1}{-4+3i} \times \frac{-4-3i}{-4-3i} = \frac{-4-3i}{(-4)^2 - (3i)^2} = \frac{-4-3i}{16 - 9i^2} = \frac{-4-3i}{16 - 9(-1)} = \frac{-4-3i}{16+9} = \frac{-4-3i}{25} = -\frac{4}{25} - \frac{3}{25}i \)

**3. Find \( \frac{z_1}{z_2} \):**
\( \frac{z_1}{z_2} = \frac{2-i}{-4+3i} \)
Multiply by the conjugate of the denominator \( (-4-3i) \):
\( \frac{2-i}{-4+3i} \times \frac{-4-3i}{-4-3i} = \frac{(2-i)(-4-3i)}{(-4+3i)(-4-3i)} \)
Numerator: \( (2-i)(-4-3i) = -8 - 6i + 4i + 3i^2 = -8 - 2i + 3(-1) = -8 - 2i - 3 = -11 - 2i \)
Denominator: \( (-4+3i)(-4-3i) = (-4)^2 - (3i)^2 = 16 - 9i^2 = 16 - 9(-1) = 16 + 9 = 25 \)
So, \( \frac{z_1}{z_2} = \frac{-11-2i}{25} = -\frac{11}{25} - \frac{2}{25}i \).
Working with complex numbers involves using conjugates for division and inverses, which helps to transform expressions into the standard rectangular form.
In simple words: For the inverse of a complex number, write 1 divided by that number, then multiply the top and bottom by its conjugate. For the division of two complex numbers, multiply the top and bottom by the conjugate of the bottom number. Simplify everything to get the final answers.

🎯 Exam Tip: Clearly show the steps for multiplying by the conjugate in both the numerator and denominator for inverse and division problems. Mistakes often happen in expanding the terms correctly.

Question 4. The complex numbers u, v, and w are related by \( \frac {1}{u} = \frac {1}{v} + \frac {1}{w} \) and \( v = 3 - 4i \), \( w = 4 + 3i \), find u in rectangular form.
Answer:
Given the relation \( \frac {1}{u} = \frac {1}{v} + \frac {1}{w} \), with \( v = 3 - 4i \) and \( w = 4 + 3i \). We need to find \( u \) in rectangular form.
First, substitute \( v \) and \( w \) into the relation:
\( \frac {1}{u} = \frac {1}{3-4i} + \frac {1}{4+3i} \)
Next, simplify each fraction by multiplying by its conjugate:
For \( \frac {1}{3-4i} \):
\( \frac {1}{3-4i} \times \frac {3+4i}{3+4i} = \frac {3+4i}{3^2 - (4i)^2} = \frac {3+4i}{9 - 16i^2} = \frac {3+4i}{9+16} = \frac {3+4i}{25} \)
For \( \frac {1}{4+3i} \):
\( \frac {1}{4+3i} \times \frac {4-3i}{4-3i} = \frac {4-3i}{4^2 - (3i)^2} = \frac {4-3i}{16 - 9i^2} = \frac {4-3i}{16+9} = \frac {4-3i}{25} \)
Now, add these two results to find \( \frac {1}{u} \):
\( \frac {1}{u} = \frac {3+4i}{25} + \frac {4-3i}{25} = \frac {(3+4i) + (4-3i)}{25} = \frac { (3+4) + (4-3)i }{25} = \frac {7+i}{25} \)
So, \( \frac {1}{u} = \frac {7+i}{25} \). To find \( u \), we take the reciprocal of this:
\( u = \frac {25}{7+i} \)
Multiply by the conjugate of \( 7+i \), which is \( 7-i \):
\( u = \frac {25}{7+i} \times \frac {7-i}{7-i} = \frac {25(7-i)}{7^2 - i^2} = \frac {25(7-i)}{49 - (-1)} = \frac {25(7-i)}{50} \)
Simplify the fraction:
\( u = \frac {7-i}{2} = \frac{7}{2} - \frac{1}{2}i \). This problem shows how to manipulate complex number fractions and reciprocals to solve for an unknown.
In simple words: First, find 1/v and 1/w by multiplying each by its conjugate. Add these two results to get 1/u. Finally, take the reciprocal of 1/u and multiply by its conjugate to find u in the simple 'a + bi' form.

🎯 Exam Tip: Break down complex problems into smaller, manageable steps. Simplify each fractional term individually before performing addition or finding the reciprocal, and always keep track of \( i^2 = -1 \).

Question 5. Prove the following properties:
(i) \( z \) is real if and only if \( z = \overline {z} \)
Answer:
(i) We need to prove that \( z \) is real if and only if \( z = \overline {z} \).
Let \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part. The conjugate of \( z \) is \( \overline{z} = x - iy \).

**Part 1: Assume \( z \) is real, then show \( z = \overline {z} \).**
If \( z \) is a real number, then its imaginary part \( y \) must be 0.
So, \( z = x + 0i = x \).
And its conjugate \( \overline{z} = x - 0i = x \).
Therefore, if \( z \) is real, then \( z = \overline {z} \).

**Part 2: Assume \( z = \overline {z} \), then show \( z \) is real.**
If \( z = \overline {z} \), then \( x + iy = x - iy \).
Subtract \( x \) from both sides: \( iy = -iy \).
Add \( iy \) to both sides: \( iy + iy = 0 \implies 2iy = 0 \).
Since \( 2i \ne 0 \), it must be that \( y = 0 \).
If \( y = 0 \), then \( z = x + 0i = x \), which means \( z \) is a real number. This property is fundamental to understanding complex numbers.
Thus, \( z \) is real if and only if \( z = \overline {z} \).
In simple words: If a number is real, its imaginary part is zero, so it is the same as its conjugate. If a number is the same as its conjugate, it means its imaginary part must be zero, making it a real number.

🎯 Exam Tip: "If and only if" proofs require showing both directions: "if A then B" and "if B then A". Clearly define your variables (like \( z = x+iy \)) at the start.

(ii) \( Re(z) = \frac{z+\overline{z}}{2} \) and \( Im(z) = \frac{z-\overline{z}}{2i} \)
Answer:
(ii) We need to prove \( Re(z) = \frac{z+\overline{z}}{2} \) and \( Im(z) = \frac{z-\overline{z}}{2i} \).
Let \( z = x + iy \), where \( x \) is the real part \( Re(z) \) and \( y \) is the imaginary part \( Im(z) \).
The conjugate of \( z \) is \( \overline{z} = x - iy \).

**Proof for \( Re(z) \):**
Consider the sum \( z + \overline{z} \):
\( z + \overline{z} = (x+iy) + (x-iy) \)
\( = x + iy + x - iy \)
\( = 2x \)
Now, divide by 2:
\( \frac{z+\overline{z}}{2} = \frac{2x}{2} = x \)
Since \( x = Re(z) \), we have proven that \( Re(z) = \frac{z+\overline{z}}{2} \).

**Proof for \( Im(z) \):**
Consider the difference \( z - \overline{z} \):
\( z - \overline{z} = (x+iy) - (x-iy) \)
\( = x + iy - x + iy \)
\( = 2iy \)
Now, divide by \( 2i \):
\( \frac{z-\overline{z}}{2i} = \frac{2iy}{2i} = y \)
Since \( y = Im(z) \), we have proven that \( Im(z) = \frac{z-\overline{z}}{2i} \). These formulas provide a useful way to extract the real and imaginary parts of a complex number using its conjugate.
In simple words: To find the real part, add the complex number and its conjugate, then divide by 2. To find the imaginary part, subtract the conjugate from the complex number, then divide by 2i.

🎯 Exam Tip: These formulas are very useful identities. Knowing them saves time in calculations and provides an alternative way to extract real and imaginary components without explicitly writing \( x+iy \).

Question 6. Find the least value of the positive integer n for which \( (\sqrt{3} + i)^n \) (i) real, (ii) purely imaginary.
Answer:
Let \( z = \sqrt{3} + i \). First, convert \( z \) to polar form \( r(\cos\theta + i\sin\theta) \).
\( r = |z| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3+1} = \sqrt{4} = 2 \)
To find \( \theta \):
\( \cos\theta = \frac{\sqrt{3}}{2} \) and \( \sin\theta = \frac{1}{2} \). This means \( \theta = \frac{\pi}{6} \) (or 30 degrees).
So, \( z = 2\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) \).
Now, using De Moivre's Theorem, \( z^n = (\sqrt{3} + i)^n = 2^n\left(\cos\left(\frac{n\pi}{6}\right) + i\sin\left(\frac{n\pi}{6}\right)\right) \).

(i) For \( (\sqrt{3} + i)^n \) to be purely real, its imaginary part must be zero.
\( \sin\left(\frac{n\pi}{6}\right) = 0 \)
This happens when \( \frac{n\pi}{6} \) is an integer multiple of \( \pi \), i.e., \( \frac{n\pi}{6} = k\pi \) for some integer \( k \).
\( \implies n = 6k \)
The least positive integer value for \( n \) occurs when \( k=1 \), so \( n = 6 \).
Let's verify for \( n=6 \):
\( (\sqrt{3} + i)^6 = 2^6 \left(\cos\left(\frac{6\pi}{6}\right) + i\sin\left(\frac{6\pi}{6}\right)\right) = 64 (\cos(\pi) + i\sin(\pi)) = 64(-1 + 0i) = -64 \). This is a real number.

(ii) For \( (\sqrt{3} + i)^n \) to be purely imaginary, its real part must be zero.
\( \cos\left(\frac{n\pi}{6}\right) = 0 \)
This happens when \( \frac{n\pi}{6} \) is an odd multiple of \( \frac{\pi}{2} \), i.e., \( \frac{n\pi}{6} = (2k+1)\frac{\pi}{2} \) for some integer \( k \).
\( \implies n = 3(2k+1) \)
The least positive integer value for \( n \) occurs when \( k=0 \), so \( n = 3(2(0)+1) = 3 \).
Let's verify for \( n=3 \):
\( (\sqrt{3} + i)^3 = 2^3 \left(\cos\left(\frac{3\pi}{6}\right) + i\sin\left(\frac{3\pi}{6}\right)\right) = 8 \left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right) = 8(0 + 1i) = 8i \). This is a purely imaginary number.
Using De Moivre's theorem greatly simplifies finding powers of complex numbers and their nature. This theorem helps convert powers into a clearer trigonometric form.
In simple words: Convert the complex number to polar form. Use De Moivre's Theorem for \( (\sqrt{3} + i)^n \). For it to be real, the sine part must be zero, which happens for \( n=6 \). For it to be purely imaginary, the cosine part must be zero, which happens for \( n=3 \).

🎯 Exam Tip: For problems involving powers of complex numbers, converting to polar form and applying De Moivre's Theorem is usually the most efficient method to determine their properties (real, imaginary, etc.).

Question 7. Show that
(i) \( (2 + i\sqrt{3})^{10} – (2 – i\sqrt{3})^{10} \) is purely imaginary.
Answer:
(i) Let \( z = (2 + i\sqrt{3})^{10} – (2 – i\sqrt{3})^{10} \).
We need to show that \( z \) is purely imaginary. A complex number is purely imaginary if \( Re(z) = 0 \) or if \( z = -\overline{z} \).
Let \( w = (2 + i\sqrt{3})^{10} \). Then \( \overline{w} = \overline{(2 + i\sqrt{3})^{10}} = (\overline{2 + i\sqrt{3}})^{10} = (2 - i\sqrt{3})^{10} \).
So, the expression can be written as \( z = w - \overline{w} \).
We know that for any complex number \( w \), \( w - \overline{w} = (x+iy) - (x-iy) = 2iy \), which is purely imaginary.
Alternatively, we can take the conjugate of \( z \):
\( \overline{z} = \overline{(2 + i\sqrt{3})^{10} – (2 – i\sqrt{3})^{10}} \)
Using the property \( \overline{z_1 - z_2} = \overline{z_1} - \overline{z_2} \):
\( \overline{z} = \overline{(2 + i\sqrt{3})^{10}} - \overline{(2 – i\sqrt{3})^{10}} \)
Using the property \( \overline{z^n} = (\overline{z})^n \):
\( \overline{z} = ( \overline{2 + i\sqrt{3}} )^{10} - ( \overline{2 – i\sqrt{3}} )^{10} \)
\( \overline{z} = (2 - i\sqrt{3})^{10} - (2 + i\sqrt{3})^{10} \)
Notice that this is \( -( (2 + i\sqrt{3})^{10} - (2 - i\sqrt{3})^{10} ) \).
So, \( \overline{z} = -z \).
Since \( \overline{z} = -z \), it means that \( z \) is a purely imaginary number. This property simplifies the process of proving the nature of complex expressions.
In simple words: Let the first part of the expression be 'w'. Then the second part is the conjugate of 'w'. The difference between a complex number and its conjugate (w - w-bar) is always a purely imaginary number. This proves the statement.

🎯 Exam Tip: Recognize patterns like \( z - \overline{z} \) and \( z + \overline{z} \). The property \( \overline{z} = -z \) is a direct indicator of a purely imaginary number (provided \( z \ne 0 \)).

(ii) \( \left(\frac{19-7i}{9+i}\right)^{12} + \left(\frac{20-5i}{7-6i}\right)^{12} \) is real.
Answer:
(ii) We need to show that \( \left(\frac{19-7i}{9+i}\right)^{12} + \left(\frac{20-5i}{7-6i}\right)^{12} \) is real.
A complex number \( Z \) is real if \( Z = \overline{Z} \). Let the given expression be \( Z \).
\( Z = A^{12} + B^{12} \), where \( A = \frac{19-7i}{9+i} \) and \( B = \frac{20-5i}{7-6i} \).

First, simplify \( A = \frac{19-7i}{9+i} \):
Multiply numerator and denominator by \( 9-i \):
\( A = \frac{19-7i}{9+i} \times \frac{9-i}{9-i} = \frac{(19-7i)(9-i)}{9^2 - i^2} = \frac{171 - 19i - 63i + 7i^2}{81+1} \)
\( A = \frac{171 - 82i - 7}{82} = \frac{164 - 82i}{82} = \frac{82(2-i)}{82} = 2-i \)

Next, simplify \( B = \frac{20-5i}{7-6i} \):
Multiply numerator and denominator by \( 7+6i \):
\( B = \frac{20-5i}{7-6i} \times \frac{7+6i}{7+6i} = \frac{(20-5i)(7+6i)}{7^2 - (6i)^2} = \frac{140 + 120i - 35i - 30i^2}{49+36} \)
\( B = \frac{140 + 85i + 30}{85} = \frac{170 + 85i}{85} = \frac{85(2+i)}{85} = 2+i \)

Now substitute \( A = 2-i \) and \( B = 2+i \) back into the expression for \( Z \):
\( Z = (2-i)^{12} + (2+i)^{12} \)
We need to show that \( Z = \overline{Z} \).
\( \overline{Z} = \overline{(2-i)^{12} + (2+i)^{12}} \)
\( \overline{Z} = \overline{(2-i)^{12}} + \overline{(2+i)^{12}} \)
\( \overline{Z} = (\overline{2-i})^{12} + (\overline{2+i})^{12} \)
\( \overline{Z} = (2+i)^{12} + (2-i)^{12} \)
\( \overline{Z} = Z \)
Since \( Z = \overline{Z} \), the expression is a real number. This problem illustrates how to simplify complex fractions before applying power properties.
In simple words: Simplify each fraction separately by multiplying the top and bottom by the conjugate of the denominator. Once simplified, you'll see that the two resulting complex numbers are conjugates of each other. When you add a number raised to a power and its conjugate raised to the same power, the result is always a real number.

🎯 Exam Tip: When proving an expression is real, showing that the expression is equal to its own conjugate \( (Z = \overline{Z}) \) is a robust and often simpler method than trying to simplify the entire expression to \( a+0i \).