Inter 1st Year Maths Complex Numbers and Quadratic Equations Solutions Exercise 4b

Practicing the AP Board Solutions Class 11 Maths and Chapter 4 Inter 1st Year Maths Complex Numbers and Quadratic Equations Solutions Exercise 4b Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Complex Numbers and Quadratic Equations Solutions Exercise 4b

Complex Numbers and Quadratic Equations Exercise 4b Solutions

Complex Numbers and Quadratic Equations Class 11 Exercise 4b Solutions – Complex Numbers and Quadratic Equations 4b Exercise Solutions

I.

Question 1.
Evaluate \(\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3\)
Solution:

= [-1 – i]³
= (-1)³ [1 + i]³
= -[1³ + i³ + 3.1.i (1 + i)]
= -[1 + i³ + 3i + 3i²]
= -[1 – i + 3i – 3]
= -[-2 + 2i]
= 2 – 2i

Question 2.
For any two complex numbers z₁ and z₂, prove that Re (z₁ z₂) = Re z₁ Re z₂ – Im z₁ Im z₂.
Solution:
Let’s assume z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ as two complex numbers
Product of these complex numbers
z₁ z₂ = (x₁ + iy₁) (x₂ + iy₂)
= x₁ (x₂ + iy₂) + iy₂ (x₂ + iy₂)
= x₁ x₂ + i x₁ y₂ + i y₁ x₂ + i² y₁ y₂ [∵ i² = -1]
= x₁ x₂ + i x₁ y₂ + i y₁ x₂ – y₁ y₂
= (x₁ x₂ – y₁ y₂) + i(x₁ y₂ + y₁ x₂)
Now, Re (z₁ z₂) = x₁ x₂ – y₁ y₂
Re (z₁ z₂) = Re z₁ Re z₂ – Im z₁ Im z₂
Hence proved.

Question 3.
Reduce \(\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)\) to the standard form.
Solution:


Hence, this is the required standard form.

Question 4.
Let z₁ = 2 – i, z₂ = -2 + i. Find
(i) \({Re}\left(\frac{Z_1 Z_2}{\bar{Z}_1}\right)\)
(ii) \({Im}\left(\frac{1}{z_1 \bar{z}_1}\right)\)
Solution:
Given z₁ = 2 – i, z₂ = -2 + i
(i) z₁ z₂ = (2 – i) (-2 + i)
= -4 + 2i + 2i – i²
= -4 + 4i – (-1)
= -3 + 4i
\(\overline{\mathrm{z}_1}\) = 2 + i
∴ \(\frac{z_1 z_2}{z_1}=\frac{-3+4 i}{2+i}\)
On multiplying the numerator and denominator by (2 – i), we get

Question 5.
Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of -6 – 24i.
Solution:
Let’s assume z = (x – iy) (3 + 5i)
= 3x + 5xi – 3yi – 5yi²
= 3x + 5xi – 3yi + 5y
= (3x + 5y) + i(5x – 3y)
∴ \(\overline{\mathrm{z}}\) = (3x + 5y) + i(5x – 3y)
Also given, \(\overline{\mathrm{z}}\) = -6 – 24i
and, (3x + 5y) + i(5x – 3y) = -6 – 24i
On equating real and imaginary parts, we have
3x + 5y = -6 ……….. (i)
5x – 3y = 24 ……….. (ii)
Performing (i) × 3 + (ii) × 5, we get
(9x + 15y) + (25x – 15y) = -18 + 120
⇒ 34x = 102
⇒ x = 3
Putting the value of x in equation (i), we get
3(3) + 5y = -6
⇒ 5y = -6 – 9 = -15
⇒ y = -3
Therefore, the values of x and y are 3 and -3, respectively.

Question 6.
Find the modulus of \(\frac{1+i}{1-i}-\frac{1-i}{1+i}\).
Solution:

Question 7.
Find the number of non-zero integral solutions of the equation |1 – i|^x = 2^x.
Solution:
|1 – i|^x = 2^x
⇒ \(\left(\sqrt{1^2+(-1)^2}\right)=2^x\)
⇒ \((\sqrt{2})^x=2^x\)
⇒ \(2^{\frac{x}{2}}=2^x\)
⇒ \(\frac {x}{2}\) = x
⇒ x = 2x
⇒ 2x – x = 0
⇒ x = 0
Therefore, 0 is the only integral solution of the given equation.

II.

Question 1.
If x – iy = \(\sqrt{\frac{a-i b}{c-i d}}\), prove that \(\left(x^2+y^2\right)^2=\frac{a^2+b^2}{c^2+d^2}\).
Solution:

Question 2.
If z₁ = 2 – i, z₂ = 1 + i, find \(\left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|\).
Solution:
Given, z₁ = 2 – i, z₂ = 1 + i

Question 3.
If a + ib = \(\frac{(x+i)^2}{2 x^2+1}\), prove that a² + b² = \(\frac{\left(x^2+1\right)^2}{\left(2 x^2+1\right)^2}\).
Solution:

Question 4.
If (x + iy)³ = u + iv, then show that \(\frac{u}{x}+\frac{v}{y}\) = 4(x² – y²).
Solution:
Given, (x + iy)³ = u + iv
⇒ x³ + (iy)³ + 3xiy (x + iy) = u + iv
⇒ x³ + i³ y³ + 3x²yi + 3x y² i² = u + iv
⇒ x³ – i y³ + 3x²yi – 3xy² = u + iv
⇒ (x³ – 3xy²) + i(3x²y – y³) = u + iv
On equating real and imaginary parts, we get
u = x³ – 3xy², v = 3x²y – y³
\(\frac{u}{x}+\frac{v}{y}=\frac{x^3-3 x y^2}{x}+\frac{3 x^2 y-y^3}{y}\)
= \(\frac{x\left(x^2-3 y^2\right)}{x}+\frac{y\left(3 x^2-y^2\right)}{y}\)
= x³ – 3y² + 3x² – y²
= 4x² – 4y²
= 4(x² – y²)
∴ \(\frac{u}{x}+\frac{v}{y}\) = 4(x² – y²)
Hence proved.

Question 5.
If α and β are different complex , numbers with |β| = 1, then find \(\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|\).
Solution:
Let α = a + ib and β = x + iy
Given, |β| = 1
So, \(\sqrt{x^2+y^2}\) = 1
⇒ x² + y² = 1 ……….(i)

Question 6.
If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²
Solution:
Given, (a + ib) (c + id) (e + if) (g + ih) = A + iB [∵ |(a + ib) (c + id) (e + if) (g + ih)| = |A + B|]
⇒ |(a + ib)| × |(c + id)| × |(e + if)| × |(g + ih)| = |A + iB|
\(\sqrt{a^2+b^2} \times \sqrt{c^2+d^2} \times \sqrt{e^2+f^2} \times \sqrt{g^2+h^2}\) = \(\sqrt{A^2+B^2}\)
On squaring both sides, we get
(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²
Hence proved.

Question 7.
If \(\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{m}}\) = 1, then find the least positive integral value of m.
Solution:

i^m = 1
Hence, m = 4K, where k is some integer.
Thus, the least positive integer is 1.
Therefore, the least positive integral value 0 is m (4 × 1) = 4

Question 8.
Show that the four points in the Argand plane represented by the complex numbers 2 + i, 4 + 3i, 2 + 5i, and 3i are the vertices of a square.
Solution:
We are given four complex numbers
z₁ = 2 + i, z₂ = 4 + 3i, z₃ = 2 + 5i, z₄ = 0 + 3i = 3i
A = z₁ = (2, 1)
B = z₂ = (4, 3)
C = z₃ = (2, 5)
D = z₄ = (0, 3)

∴ Diagonals are equal to 4
Vector AC = C – A
= (2, 5) – (2, 1)
= (0, 4)
Vector BD = D – B
= (0, 3) – (4, 3)
= (-4, 0)
Dot product = (0, 4) . (-4, 0)
= 0 . (-4) + 4 . 0
= 0
Diagonals are perpendicular.
Hence, the given complex numbers represented the vertices of a square in the Argand plane.
We are given four complex numbers.

Question 9.
Show that the points in the Argand plane represented by the complex numbers -2 + 7i, \(\frac{-3}{2}+\frac{1}{2} i\), 4 – 3i, \(\frac {7}{2}\)(1 + i) are the vertices of a rhombus.
Solution:
We are given four complex numbers


All four sides have a length of √42.5
Diagonal AC:
C – A = (4 – (-2), -3 – 7) = (6, -10)
Diagonal BD:
D – B = \(\left(\frac{7}{2}+\frac{3}{2}, \frac{7}{2}-\frac{1}{2}\right)\) = (5, 3)
Dot product = \(\overline{\mathrm{AC}} \cdot \overline{\mathrm{BD}} .\)
= (6) (5) + (-10) (3)
= 30 – 30
= 0
Diagonals are perpendicular.
It’s a rhombus.
Therefore, the points represented by the complex numbers from a rhombus in the Argand plane.

Question 10.
The points P and Q denote the complex numbers z₁, z₂ in the Argand diagram. O is the origin. If \(\overline{z_1} \overline{z_2}+\overline{z_1} z_2\) = 0 then show that \(\mathrm{POQ}=\frac{\pi}{2}\)
Solution:
For complex numbers

Question 11.
If the complex number z has agrument θ, 0 < θ < \(\frac{\pi}{2}\) and satisfy the equation |z – 3i| = 3. Then prove that (cot θ – \(\frac {6}{z}\)) = i.
Solution:
Let z = x + iy
|z – 3i| = |x + i(y – 3)|
⇒ \(\sqrt{x^2+(y-3)^2}\) = 3
⇒ x² + (y – 3)² = 9
⇒ (x² + y² – 6y + 9) = 9
⇒ x² + y² = 6y ………(1)
Since z = x + iy, we have agrument (z) = θ
⇒ \(\tan ^{-1}\left(\frac{y}{x}\right)\) = θ
⇒ cot θ = \(\frac {x}{y}\) ……….(2)
A complex number z, such that argument (z) = θ, where