Class 11 - Mathematics
Complex Numbers - Exercise 5 - Misc
Top Block 1
Question 1:
Evaluate: [i18 + (1/i)25]3
Answer:
Given, [i18 + (1/i)25]3
= [i4*4+2 + 1/i4*6+1]3
= [(i4)2 * i2 + 1/{(i4)6 * i}]3
= [-1 + 1/i]3 [Since i4 = 1 and i2 = -1]
= [-1 + i/i2]3
= [-1 – i]3
= (-1)3 * [1 + i]3
= – {1 + i3 + 3 * 1 * i(1 + i)}
= – {1 – i + 3i + 3i2} [Since i2 = -1]
= – {1 – i + 3i – 3}
= -(-2 + 2i)
= 2 – 2i
Question 2:
For any two complex numbers z1 and z2, prove that
Re (z1z2) = Re z1 Re z2 – Im z1 Im z2
Answer:
Let z1 = x1 + iy1 and z2 = x2 + iy2
Now, z1z2 = (x1 + iy1)(x2 + iy2)
= x1(x2 + iy2) + iy1(x2 + iy2)
= x1x2 + i x1y2 + iy1x2 + i2 y2y2
= x1x2 + i x1y2 + iy1x2 – y2y2 [Since i2 = -1]
= (x1x2 – y1y2) + i(x1y2 + y1x2)
So, Re (z1z2) = x1x2 – y1y2
⇒ Re (z1z2) = Re z1 Re z2 – Im z1 Im z2
Hence, proved.
Question 3:
Reduce {1/(1 – 4i) – 2/(1 + i)}{(3 – 4i)/(5 + i)} to the standard form.
Answer:
Given, {1/(1 – 4i) – 2/(1 + i)}{(3 – 4i)/(5 + i)}
= [{(1 + i) – 2(1 – 4i)}/{(1 – 4i)(1 + i)}]{(3 – 4i)/(5 + i)}
= [{1 + i – 2 + 8i}/{1 + i – 4i – 4i2}]{(3 – 4i)/(5 + i)}
= [{-1 + 9i}/{1 + i – 4i + 4}]{(3 – 4i)/(5 + i)}
= [{-1 + 9i}/{5 – 3i}]{(3 – 4i)/(5 + i)}
= [{-3 + 4i + 27i – 36 i2}/{25 + 5i – 15i – 3i2}]
= [{-3 + 4i + 27i + 36}/{25 + 5i – 15i + 3}]
= [{-3 + 4i + 27i + 36}/{25 + 5i – 15i + 3}]
= (33 + 31i)/(28 – 10i)
= (33 + 31i)/{2 * (14 – 5i)}
= [(33 + 31i)/{2 * (14 – 5i)}] *{2 * (14 + 5i)/ (14 + 5i)}
= [{462 + 165i + 434i + 155 i2}/[2 * (142 – (5i)2]
= [{462 + 165i + 434i – 155}/{2 * (196 + 25)}
= (307 + 599i)/(2 * 221)
= (307 + 599i)/442
= 307/442 + 599i/442
This is the required standard form.
Question 4:
If x – iy = √{(a – ib)/(c – id)}, prove that (x2 + y2)2 = (a2 + b2)/(c2 + d2)
Answer:
Given, x – iy = √{(a – ib)/(c – id)}
= √[{(a – ib)/(c – id)} * {(c + id)/(c + id)}]
= √[{(ac + bd) + i(ad – bc)}/(c2 + d2)]
Now, (x – iy)2 = {(ac + bd) + i(ad – bc)}/(c2 + d2)
⇒ x2 – y2 – 2ixy = {(ac + bd) + i(ad – bc)}/(c2 + d2)
⇒ (x2 – y2) – 2ixy = (ac + bd)/(c2 + d2) + i(ad – bc)/(c2 + d2)
On comparing real and imaginary part, we get
x2 – y2 = (ac + bd)/(c2 + d2) and -2xy = (ad – bc)/(c2 + d2) ………………1
Now, (x2 + y2)2 = (x2 – y2)2 + 4 x2y2
= {(ac + bd)/(c2 + d2)}2 + {(ad – bc)/(c2 + d2)}2
= (a2c2 + b2d2 + 2abcd + a2d2 + b2c2 – 2abcd)/(c2 + d2)2
= (a2c2 + b2d2 + a2d2 + b2c2)/(c2 + d2)2
= {a2(c2 + d2) + b2(c2 + d2)}/(c2 + d2)2
= {(a2 + b2)(c2 + d2)}/(c2 + d2)2
= (a2 + b2)/(c2 + d2)
Hence, proved.
Question 5:
Convert the following in the polar form:
(i) (1 + 7i)/(2 – i)2 (ii) (1 + 3i)/(1 – 2i)
Answer:
(i) Let z = (1 + 7i)/(2 – i)2
= (1 + 7i)/(4 + i2 – 4i)
= (1 + 7i)/(4 – 1 – 4i)
= (1 + 7i)/(3 – 4i)
= {(1 + 7i)/(3 – 4i)} * {(3 + 4i)/ (3 + 4i)}
= (3 + 4i + 21i + 28i2)/{32 – (4i)2}
= (3 + 4i + 21i – 28)/(9 + 16)
= (-25 + 25i)/25
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2(cos2 θ + sin2 θ) = 1 + 1
⇒ r2 = 2 [Since cos2 θ + sin2 θ = 1]
⇒ r = √2 [Since r > 0]
Now, √2 cos θ = -1 and √2 sin θ = 1
⇒ cos θ = -1/√2 and sin θ = 1/√2
⇒ θ = π – π/4 = 3π/4
Now, z = r cos θ + i r sin θ
= √2(cos 3π/4 + i sin 3π/4)
This is the required polar form.
(ii) Let z = (1 + 3i)/(1 – 2i)
= {(1 + 3i)/(1 – 2i)} * {(1 + 2i)/(1 + 2i)}
= (1 + 2i + 3i + 6i2)/{12 – (2i)2}
= (1 + 2i + 3i – 6)/(1 + 4)
= (-5 + 5i)/5
= -1 + i
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2(cos2 θ + sin2 θ) = 1 + 1
⇒ r2 = 2 [Since cos2 θ + sin2 θ = 1]
⇒ r = √2 [Since r > 0]
Now, √2 cos θ = -1 and √2 sin θ = 1
⇒ cos θ = -1/√2 and sin θ = 1/√2
⇒ θ = π – π/4 = 3π/4
Now, z = r cos θ + i r sin θ
= √2(cos 3π/4 + i sin 3π/4)
This is the required polar form.
Question 6:
Solve the equation 3x2 – 4x + 20/3 = 0
Answer:
The given quadratic equation is 3x2 – 4x + 20/3 = 0
This equation can also be written as 9x2 – 16x + 20 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 9, b = –12, and c = 20
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–12)2 – 4 * 9 * 20 = 144 – 720 = –576
Therefore, the required solutions are = (-b ± √D)/2a
= {12 ± √(-576)}/(2 * 9)
= {12 ± √(-1) * √576}/18
= (12 ± i24)/18 [Since i = √(-1)]
= 6(2 ± i4)/18
= (2 ± i4)/3
Question 7:
Solve the equation x2 – 2x + 3/2 = 0
Answer:
The given quadratic equation is x2 – 2x + 3/2 = 0
This equation can also be written as 2x2 – 4x + 3 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = –4, and c = 3
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–4)2 – 4 * 2 * 3 = 16 – 24 = –8
Therefore, the required solutions are = (-b ± √D)/2a
= {4 ± √(-8)}/(2 * 2)
= {4 ± √(-1) * √8}/4
= (4 ± i2√2)/4 [Since i = √(-1)]
= 2(2 ± i√2)/4 = (2 ± i√2)/2
Question 8:
Solve the equation 27x2 – 10x + 1 = 0
Answer:
The given quadratic equation is 27x2 – 10x + 1 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 27, b = –10, and c = 1
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–10)2 – 4 * 27 * 1 = 100 – 108 = –8
Therefore, the required solutions are = (-b ± √D)/2a
= {10 ± √(-8)}/(2 * 27)
= {10 ± √(-1) * √8}/54
= (10 ± i2√2)/54 [Since i = √(-1)]
= 2(5 ± i√2)/54
= (5 ± i√2)/27
Question 9:
Solve the equation 21x2 – 28x + 10 = 0
Answer:
The given quadratic equation is 28x2 – 28x + 10 = 0
On comparing this equation with ax2 + bx + c = 0, we get
a = 28, b = –28, and c = 10
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–28)2 – 4 * 21 * 10 = 784 – 840 = –56
Therefore, the required solutions are = (-b ± √D)/2a
= {28 ± √(-56)}/(2 * 21)
= {28 ± √(-1) * √56}/42
= (28 ± i2√14)/42 [Since i = √(-1)]
= 2(14 ± i√14)/42
= (14 ± i√14)/21
= 14/21 ± i√14/21
= 2/3 ± i√14/21
Question 10:
If z1 = 2 – i, z2 = 1 + i, find |z1 + z2 + 1|/|z1 – z2 + i|
Answer:
Given, z1 = 2 – i, z2 = 1 + i
Now, |z1 + z2 + 1|/|z1 – z2 + i| =|2 – i + 1 + i + 1|/|2 – i – 1 – i + i|
=|4/(2 – 2i)|
=|4/{2(1 – i)|
= |2/(1 – i)|
= |{2/(1 – i)} * {(1 + i)/ (1 + i)}|
= |2(1 + i)/(12 – i2)|
= |2(1 + i)/(1 + 1)|
= |2(1 + i)/2|
= |(1 + i)|
= √(12 + 12)
= √2
Thus, the value of |z1 + z2 + 1|/|z1 – z2 + i| = √2
Mddle block 1
Question 11:
If a + ib = (x + i)2/(2x2 + 1), prove that a2 + b2 = (x2 + 1)2/(2x + 1)2
Answer:
Given, a + ib = (x + i)2/(2x2 + 1)
= (x2 + i2 + 2ix)/(2x2 + 1)
= (x2 – 1 + 2ix)/(2x2 + 1)
= (x2 – 1)/ (2x2 + 1) + 2ix/(2x2 + 1)
On comparing real and imaginary parts, we get
a = (x2 – 1)/ (2x2 + 1), b = 2x/(2x2 + 1)
Now, a2 + b2 = [(x2 – 1)/ (2x2 + 1)]2 + [2x/(2x2 + 1)]2
= (x4 + 1 – 2x2 + 4x2)/(2x2 + 1)2
= (x4 + 1 + 2x2)/(2x2 + 1)2
= (x2 + 1)2/(2x2 + 1)2
So, a2 + b2 = (x2 + 1)2/(2x2 + 1)2
Hence, proved.
Question 12:
Let z1 = 2 – i, z2 = -2 + i. Find
(i) Re{(z1z2)/z1} (ii) Im(1/z1z1)
Answer:
Given, z1 = 2 – i, z2 = -2 + i
(i) z1z2 = (2 – i)(-2 + i)
= -4 + 2i + 2i – i2
= -4 + 4i – (-1)
= -4 + 4i + 1
= -3 + 4i
z1 = 2 + i
So, (z1z2)/z1 = (-3 + 4i)/(2 + i)
= {(-3 + 4i)/(2 + i)} * {(2 – i)/ (2 – i)}
= (-6 + 3i + 8i – 4i2)/(22 + 12)
= (-6 + 3i + 8i + 4)/5
= (-2 + 11i)/5
= -2/5 + 11i/5
On comparing real part, we get
Re{(z1z2)/z1} = -2/5
(ii) (1/z1z1) = 1/{(2 – i)(2 + i)} = 1/(22 + 12) = 1/5
On comparing imaginary part, we get
Im (1/z1z1) = 0
Question 13:
Find the modulus and argument of the complex number (1 + 2i)/(1 – 3i)
Answer:
Let z = (1 + 2i)/(1 – 3i)
= {(1 + 2i)/(1 – 3i)} * {(1 + 3i)/(1 + 3i)}
= (1 + 3i + 2i + 6i2)/(12 + 32)
= (1 + 3i + 2i – 6)/10
= (-5 + 5i)/10
= -1/2 + 1i/2
Given, z = r cos θ + i r sin θ
Let r cos θ = -1/2 and r sin θ = 1/2
On squaring and adding, we obtain
(r cos θ)2 + (r sin θ)2 = (-1/2)2 + (1/2)2
⇒ (r2 cos2 θ + r2 sin2 θ) = 1/4 + 1/4
⇒ r2 (cos2 θ + sin2 θ) = 1/2
⇒ r2 = 1/2 [Since cos2 θ + sin2 θ = 1]
⇒ r = 1/√2 [Since r > 0]
So, Modulus = 1/√2
Now, cos θ/√2 = -1/2 and sin θ/√2 = 1/2
⇒ cos θ = -1/√2 and sin θ = 1/√2
Now, argument θ = π – π/4 = 3π/4
Therefore, the modulus and argument of the given complex number are 1/√2 and 3π/4
respectively.
Question 14:
Find the real numbers x and y if (x – iy)(3 + 5i) is the conjugate of –6 – 24i.
Answer:
Given, (x – iy)(3 + 5i) = 3x + I * 5x – I * 3y + 5y
= (3x + 5y) + i(5x – 3y)
Conjugate of (x – iy)(3 + 5i) = (3x + 5y) – i(5x – 3y)
Now, (3x + 5y) – i(5x – 3y) = -6 – 24i
Compare real and imaginary term, we get
3x+ 5y = -6 ………..1
and 5x – 3y = 24 ………..2
After solving equation 1 and 2, we get
x =3 and y= -3
Thus, the values of x and y are 3 and –3 respectively.
Question 15:
Find the modulus of {(1 + i)/(1 – i)} – {(1 – i)/(1 + i)}
Answer:
Given expression is
{(1 + i)/(1 – i)} – {(1 – i)/(1 + i)}
= [{(1 + i)*(1 + i)}/{(1 – i)*(1 + i)}] – [{(1 – i)*(1 – i)}/{(1 + i)*(1 – i)}]
= {(1 + i)2/(1 – i2)} – {(1 – i)2/(1 – i2)}
= (1 + i2 + 2i)/(1 + 1) – (1 + i2 – 2i)/(1 + 1) (since i2 = -1)
= (1 – 1 + 2i)/2 – (1- 1 – 2i)/2
= 2i/2 + 2i/2
= i + i
= 2i
= 0 + 2i
Now modulus = √(02 + 22) = √4 = 2
So, modulus of the expression = 2
Question 16:
If (x + iy)3 = u + iv, then show that: u/x + v/y = 4(x2 – y2)
Answer:
Given, (x + iy)3 = u + iv
⇒ x3 + (iy)3 + 3 * x * iy(x + iy) = u + iv
⇒ x3 – iy3 + 3x2yi + 3xy2i2 = u + iv
⇒ x3 – iy3 + 3x2yi – 3xy2 = u + iv
⇒ (x3 – 3xy2) + i(3x2y – y3) = u + iv
On comparing real and imaginary part, we get
u = x3 – 3xy2 and v = 3x2y – y3
Now, u/x + y/v = (x3 – 3xy2)/x + (3x2y – y3)/y
= x2 – 3y2 + 3x2 – y2
= 4 x2 – 4y2
= 4(x2 – y2)
So, u/x + v/y = 4(x2 – y2)
Question 17:
If α and β are different complex numbers with |β| = 1, then find |(β – α)/(1 – αβ)|
Answer:
Let α = a + ib and β = x + iy
It is given that |β| = 1
⇒ √(x2 + y2) = 1
⇒ x2 + y2 = 1 ………..1
Now, |(β – α)/(1 – αβ)|= |{(x + iy) – (a – ib)}/{1 – (a – ib)(x + iy)}|
= |{(x – a) + i(y – b)}/{1 – (ax + aiy – ibx + by)}|
= |{(x – a) + i(y – b)}/{(1 – ax – by) + i(bx – ay)}|
= |{(x – a) + i(y – b)}|/|{(1 – ax – by) + i(bx – ay)}|
= √{(x – a)2 + (y – b)2}/√{(1 – ax – by)2 + (bx – ay)2}
= √(x2 + a2 – 2ax + y2 + b2 – 2by)
√(1 + a2x2 – b2y2 -2ax + 2abxy – 2by + b2x2 – a2y2 – 2abxy)
= √{(x2 + a2) – 2ax + y2 + b2 – 2by)}
√{1 + a2(x2 + y2) + b2 (y2 + x2) – 2ax – 2by}
= √{1 + a2 + b2 – 2ax – 2by)}/ √{1 + a2 + b2 – 2ax – 2by)} [From eq. 1]
= 1
So, |(β – α)/(1 – αβ)|= 1
Question 18:
Find the number of non-zero integral solutions of the equation |1 – i|x = 2x
Answer:
Given, |1 – i|x = 2x
⇒ √{12 + (-1)2} x = 2x
⇒ √{1 + 1} x = 2x
⇒ (√2)x = 2x
⇒ 2x/2 = 2x
⇒ x/2 = x
⇒ x = 2x
⇒ 2x – x = 0
⇒ x = 0
Thus, 0 is the only integral solution of the given equation. Therefore, the number of nonzero
integral solutions of the given equation is 0.
Question 19:
If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that:
(a2 + b2)(c2 + d2)(e2 + f2)(g2 + h2) = A2 + B2
Answer:
Given, (a + ib)(c + id)(e + if)(g + ih) = A + iB
⇒ |(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|
⇒ |(a + ib)||(c + id)||(e + if)||(g + ih)| = |A + iB|
⇒ √(a2 + b2) √(c2 + d2) √(e2 + f2) √(g2 + h2) = √(A2 + B2)
On squaring both sides, we get
(a2 + b2)(c2 + d2)(e2 + f2)(g2 + h2) = A2 + B2
Hence, proved.
Question 20:
If {(1 + i)/(1 – i)}m then find the least positive integral value of m.
Answer:
Given, {(1 + i)/(1 – i)}m = 1
⇒ [{(1 + i) * (1 + i)}/{(1 – i) * (1 + i)}]m = 1
⇒ [{(1 + i)2}/{(1 – i2)}]m = 1
⇒ [(1 + i2 + 2i)/{1 – (-1)}]m = 1 [Since i2 = -1]
⇒ [(1 – 1 + 2i)/{1 + 1}]m = 1
⇒ [2i/2]m = 1
⇒ im = 1
Now, im is 1 when n = 4 [Since i4 = 1]
So, the least value of n is 4
Bottom Block 3
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