NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

Question 1: Express each of the complex numbers in the form a+ib.

(1) (5i)(−35i)

(2) i9+i19

(3) i−39

(4) 3(7+7i)+i(7+7i)

(5) (1−i)−(−1+6i)

(6) (15+i25)−(4+i52)

(7) [(13+i73)+(4+i13)]−(−43+i)

(8) (1−i)4

(9) (13+3i)3

(10)(−2−13i)3

Answer:

(1) On solving

(5i)(−35i)
we will get

(5i)(−35i)=5×(−35)×i×i
=−3×i2 (∵i2=−1)
=−3×−1
=3

Now, in the form of a+ib, we can write it as
=3+0i

(2)

We know that i4=1
Now, we will reduce i9+i19 into

=(1)2⋅i+(1)3⋅(−i)(∵i4=1,i3=−i)2⋅i+(i4)3⋅i3=i−i=0

Now, in the form of a+ib we can write it as

o+io

Therefore, the answer is o+io

(3)

We know that i 4 = 1
Now, we will reduce i − 39 into

i − 39 = ( i 4 ) − 9 ⋅ i − 3 = ( 1 ) − 9 ⋅ ( − i ) − 1 ( ∵ i 4 = 1 , i 3 = − i ) = 1 − i = 1 − i × i i = i − i 2 ( ∵ i 2 = − 1 ) = i − ( − 1 ) = i

Now, in the form of a + i b we can write it as

o + i 1

Therefore, the answer is o + i 1

(4)

Given problem is

3 ( 7 + 7 i ) + i ( 7 + 7 i )

Now, we will reduce it into

3 ( 7 + 7 i ) + i ( 7 + 7 i ) = 21 + 21 i + 7 i + 7 i 2 = 21 + 21 i + 7 i + 7 ( − 1 ) ( ∵ i 2 = − 1 ) = 21 + 21 i + 7 i − 7 = 14 + 28 i

Therefore, the answer is 14 + i 28

(5)

( 1 − i ) − ( − 1 + 6 i )

Answer:
Given problem is

( 1 − i ) − ( − 1 + 6 i )

Now, we will reduce it into

( 1 − i ) − ( − 1 + 6 i ) = 1 − i + 1 − 6 i = 2 − 7 i

Therefore, the answer is 2 − 7 i

(6)

Given problem is

( 1 5 + i 2 5 ) − ( 4 + i 5 2 )

Now, we will reduce it into

( 1 5 + i 2 5 ) − ( 4 + i 5 2 ) = 1 5 + i 2 5 − 4 − i 5 2 = 1 − 20 5 + i ( 4 − 25 ) 10 = − 19 5 − i 21 10

Therefore, the answer is − 19 5 − i 21 10

(7)

Given problem is

[ ( 1 3 + i 7 3 ) + ( 4 + i 1 3 ) ] − ( − 4 3 + i )

Now, we will reduce it into

[ ( 1 3 + i 7 3 ) + ( 4 + i 1 3 ) ] − ( − 4 3 + i ) = 1 3 + i 7 3 + 4 + i 1 3 + 4 3 − i = 1 + 4 + 12 3 + i ( 7 + 1 − 3 ) 3 = 17 3 + i 5 3

Therefore, the answer is 17 3 + i 5 3

(8)

The given problem is

( 1 − i ) 4
Now, we will reduce it into

( 1 − i ) 4 = ( ( 1 − i ) 2 ) 2 = ( 1 2 + i 2 − 2.1 . i ) 2 ( u sing ( a − b ) 2 = a 2 + b 2 − 2 a b ) = ( 1 − 1 − 2 i ) 2 ( ∵ i 2 = − 1 ) = ( − 2 i ) 2 = 4 i 2 = − 4

Therefore, the answer is − 4 + i 0

Given problem is

( 1 3 + 3 i ) 3

Now, we will reduce it into

( 1 3 + 3 i ) 3 = ( 1 3 ) 3 + ( 3 i ) 3 + 3 ⋅ ( 1 3 ) 2 ⋅ 3 i + 3 ⋅ 1 3 ⋅ ( 3 i ) 2 ( u sing ( a + b ) 3 = a 3 + b 3 + 3 a 2 b + 3 a b 2 ) = 1 27 + 27 i 3 + i + 9 i 2 = 1 27 + 27 ( − i ) + i + 9 ( − 1 ) ( ∵ i 3 = − i and i 2 = − 1 ) = 1 27 − 27 i + i − 9 = 1 − 243 27 − 26 i = − 242 27 − 26 i

Therefore, the answer is

− 242 27 − 26 i
(10)

Given problem is

( − 2 − 1 3 i ) 3

Now, we will reduce it into

( − 2 − 1 3 i ) 3 = − ( ( 2 ) 3 + ( 1 3 i ) 3 + 3 ⋅ ( 2 ) 2 1 3 i + 3 ⋅ ( 1 3 i ) 2 ⋅ 2 ) ( u sing ( a + b ) 3 = a 3 + b 3 + 3 a 2 b + 3 a b 2 ) = − ( 8 + 1 27 i 3 + 3 ⋅ 4 ⋅ 1 3 i + 3 ⋅ 1 9 i 2 ⋅ 2 ) = − ( 8 + 1 27 ( − i ) + 4 i + 2 3 ( − 1 ) ) ( ∵ i 3 = − i and i 2 = − 1 ) = − ( 8 − 1 27 i + 4 i − 2 3 ) = − ( ( − 1 + 108 ) 27 i + 24 − 2 3 ) = − 22 3 − i 107 27

Therefore, the answer is − 22 3 − i 107 27

Question 11: Find the multiplicative inverse of each of the complex numbers.

4−3i

Answer:

Let z=4−3i
Then,
z¯=4+3i
And
|z|2=42+(−3)2=16+9=25
Now, the multiplicative inverse is given by
z−1=z¯|z|2=4+3i25=425+i325

Therefore, the multiplicative inverse is

425+i325

Question 12: Find the multiplicative inverse of each of the complex numbers.

5+3i

Answer:

Let z=5+3i
Then,
z¯=5−3i
And
|z|2=(5)2+(3)2=5+9=14
Now, the multiplicative inverse is given by
z−1=z¯|z|2=5−3i14=514−i314

Therefore, the multiplicative inverse is 514−i314

Question 13: Find the multiplicative inverse of each of the complex numbers.

−i

Answer:

Let z=−i
Then,
z¯=i
And
|z|2=(0)2+(1)2=0+1=1
Now, the multiplicative inverse is given by
z−1=z¯|z|2=i1=0+i

Therefore, the multiplicative inverse is 0+i1

Question 14: Express the following expression in the form of a+ib:

(3+i5)(3−i5)(3+2i)−(3−i2)

Answer:

Given problem is
(3+i5)(3−i5)(3+2i)−(3−i2)
Now, we will reduce it into

(3+i5)(3−i5)(3+2i)−(3−i2)=32−(5i)2(3+2i)−(3−i2) (using (a−b)(a+b)=a2−b2)
=9−5i23+2i−3+2i
=9−5(−1)22i (∵i2=−1)
=1422i×2i2i
=72i2i2
=−72i2
Therefore, the answer is 0−i722

Question 1: Evaluate [i18+(1i)25]3 .

Answer:

The given problem is
[i18+(1i)25]3
Now, we will reduce it into
[i18+(1i)25]3=[(i4)4.i2+1(i4)6.i]3
=[14.(−1)+116.i]3 (∵i4=1,i2=−1)
=[−1+1i]3
=[−1+1i×ii]3
=[−1+ii2]3
=[−1+i−1]3=[−1−i]3
Now,
−(1+i)3=−(13+i3+3.12.i+3.1.i2) (using (a+b)3=a3+b3+3.a2.b+3.a.b2)
=−(1−i+3i+3(−1)) (∵i3=−i,i2=−1)
=−(1−i+3i−3)=−(−2+2i)
=2−2i
Therefore, the answer is 2−2i

Question 2: For any two complex numbers z1 and z2 , prove that Re(z1z2)=Rez1Rez2−Imz1Imz2

Answer:

Let two complex numbers are
z1=x1+iy1
z2=x2+iy2
Now,
z1.z2=(x1+iy1).(x2+iy2)
=x1x2+ix1y2+iy1x2+i2y1y2
=x1x2+ix1y2+iy1x2−y1y2 (∵i2=−1)
=x1x2−y1y2+i(x1y2+y1x2)
Re(z1z2)=x1x2−y1y2
=Re(z1z2)−Im(z1z2)

Hence proved

Question 3: Reduce (11−4i−21+i)(3−4i5+i) to the standard form.

Answer:

Given problem is
(11−4i−21+i)(3−4i5+i)
Now, we will reduce it into

(11−4i−21+i)(3−4i5+i)=((1+i)−2(1−4i)(1+i)(1−4i))(3−4i5+i)
=(1+i−2+8i1−4i+i−4i2)(3−4i5+i)
=(−1+9i1−3i−4(−1))(3−4i5+i)
=(−1+9i5−3i)(3−4i5+i)
=(−3+4i+27i−36i225+5i−15i−3i2)=(−3+31i+3625−10i+3)=33+31i28−10i=33+31i2(14−5i)

Now, multiply the numerator and denominator by (14+5i)
⇒33+31i2(14−5i)×14+5i14+5i
⇒462+165i+434i+155i22(142−(5i)2) (using (a−b)(a+b)=a2−b2)
⇒462+599i−1552(196−25i2)
⇒307+599i2(196+25)=307+599i2×221=307+599i442=307442+i599442

Therefore, answer is 307442+i599442

Question 4: If x−iy=a−ibc−id , prove that (x2+y2)2=a2+b2c2+d2.

Answer:

the given problem is

x−iy=a−ibc−id
Now, multiply the numerator and denominator by

c+id
x−iy=a−ibc−id×c+idc+id
=(ac+bd)+i(ad−bc)c2−i2d2=(ac+bd)+i(ad−bc)c2+d2
Now, square both the sides
(x−iy)2=((ac+bd)+i(ad−bc)c2+d2)2
=(ac+bd)+i(ad−bc)c2+d2
x2−y2−2ixy=(ac+bd)+i(ad−bc)c2+d2
On comparing the real and imaginary parts, we obtain

x2−y2=ac+bdc2+d2 and −2xy=ad−bcc2+d2 −(i)

Now,
(x2+y2)2=(x2−y2)2+4x2y2
=(ac+bdc2+d2)2+(ad−bcc2+d2)2 (using (i))
=a2c2+b2d2+2acbd+a2d2+b2c2−2adbc(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+3i1−2i

Answer(i):

Let
z=1+7i(2−i)2=1+7i4+i2−4i=1+7i4−1−4i=1+7i3−4i

Now, multiply the numerator and denominator by 3+4i
⇒z=1+7i3−4i×3+4i3+4i=3+4i+21i+28i232+42=−25+25i25=−1+i
Now,
let
rcos⁡θ=−1 and rsin⁡θ=1
On squaring both and then add
r2(cos2⁡θ+sin2⁡θ)=(−1)2+12
r2=2
r=2 (∵r>0)
Now,
2cos⁡θ=−1 and 2sin⁡θ=1
cos⁡θ=−12 and sin⁡θ=12
Since the value of cos⁡θ is negative and sin⁡θ is positive this is the case in II quadrant
Therefore,
θ=π−π4=3π4 (lies in II quadrant)
z=rcos⁡θ+irsin⁡θ
=2cos⁡3π4+i2sin⁡3π4
=2(cos⁡3π4+isin⁡3π4)
Therefore, the required polar form is

2(cos⁡3π4+isin⁡3π4)

Answer(ii):

Let
z=1+3i1−2i

Now, multiply the numerator and denominator by 1+2i
⇒z=1+3i1−2i×1+2ii+2i=1+2i+3i−61+4=−5+5i5=−1+i
Now,
let
rcos⁡θ=−1 and rsin⁡θ=1
On squaring both and then add
r2(cos2⁡θ+sin2⁡θ)=(−1)2+12
r2=2
r=2 (∵r>0)
Now,
2cos⁡θ=−1 and 2sin⁡θ=1
cos⁡θ=−12 and sin⁡θ=12
Since the value of cos⁡θ is negative and sin⁡θ is positive this is the case in II quadrant
Therefore,
θ=π−π4=3π4 (lies in II quadrant)
z=rcos⁡θ+irsin⁡θ
=2cos⁡3π4+i2sin⁡3π4
=2(cos⁡3π4+isin⁡3π4)
Therefore, the required polar form is

2(cos⁡3π4+isin⁡3π4)

Question 6: If z1=2−i,z2=1+i , find |z1+z2+1z1−z2+1| .

Answer:

It is given that
z1=2−i,z2=1+i
Then,
|z1+z2+1z1−z2+1|=|2−i+1+i+12−i−1−i+1|=|42(1−i)|=|2(1−i)|
Now, multiply the numerator and denominator by 1+i
⇒|2(1−i)×1+i1+i|=|2(1+i)12−i2|=|2(1+i)1+1|=|1+i|
Now,
|1+i|=12+12=1+1=2
Therefore, the value of

|z1+z2+1z1−z2+1| is 2

Question 7: If a+ib=(x+i)22x2+1 , prove that a2+b2=(x2+1)2(2x2+1)2 .

Answer:

It is given that
a+ib=(x+i)22x2+1
Now, we will reduce it into

a+ib=(x+i)22x2+1=x2+i2+2xi2x2+1=x2−1+2xi2x2+1=x2−12x2+1+i2x2x2+1
On comparing real and imaginary parts. we will get
a=x2−12x2+1 and b=2x2x2+1
Now,
a2+b2=(x2−12x2+1)2+(2x2x2+1)2
=x4+1−2x2+4x2(2x2+1)2
=x4+1+2x2(2x2+1)2
=(x2+1)2(2x2+1)2
Hence proved

Question 8: Find the modulus of 1+i1−i−1−i1+i .

Answer:

Let
z=1+i1−i−1−i1+i
Now, we will reduce it into
z=1+i1−i−1−i1+i=(1+i)2−(1−i)2(1+i)(1−i)=12+i2+2i−12−i2+2i12−i2 =4i1+1=4i2=2i
Now,
rcos⁡θ=0 and rsin⁡θ=2
square and add both sides. we will get,
r2cos2⁡θ+r2sin2⁡θ=02+22
r2(cos2⁡θ+sin2⁡θ)=4
r2=4 (∵cos2⁡θ+sin2⁡θ=1)
r=2 (∵r>0)

Therefore, the modulus of

1+i1−i−1−i1+i is 2

Question 9: If (x+iy)3=u+iv , then show that ux+vy=4(x2−y2).

Answer:

it is given that
(x+iy)3=u+iv
Now, expand the Left-hand side
x3+(iy)3+3.(x)2.iy+3.x.(iy)2=u+iv
x3+i3y3+3x2iy+3xi2y2=u+iv
x3−iy3+3x2iy−3xy2=u+iv (∵i3=−i and i2=−1)
x3−3xy2+i(3x2y−y3)=u+iv
On comparing real and imaginary parts. we will get,
u=x3−3xy2 and v=3x2y−y3
Now,
ux+vy=x(x2−3y2)x+y(3x2−y2)y
=x2−3y2+3x2−y2
=4x2−4y2
=4(x2−y2)
Hence proved

Question 10: If α and β are different complex numbers with |β|=1 , then find |β−α1−α¯β| .

Answer:

Let
α=a+ib and β=x+iy.
It is given that
|β|=1⇒x2+y2=1⇒x2+y2=1
and
α¯=a−ib
Now,
|β−α1−α¯β|=|(x+iy)−(a+ib)1−(a−ib)(x+iy)|=|(x−a)+i(y−b)1−(ax+iay−ibx−i2yb)|
=|(x−a)+i(y−b)(1−ax−yb)−i(bx−ay)|
=(x−a)2+(y−b)2(1−ax−yb)2+(bx−ay)2
=x2+a2−2xa+y2+b2−yb1+a2x2+b2y2−2ax+2abxy−by+b2x2+a2y2−2abxy
=(x2+y2)+a2−2xa+b2−yb1+a2(x2+y2)+b2(x2+y2)−2ax+2abxy−by−2abxy
=1+a2−2xa+b2−yb1+a2+b2−2ax−by (∵x2+y2=1 given)
=1

Therefore, value of |β−α1−α¯β| is 1

Question 11: Find the number of non-zero integral solutions of the equation |1−i|x=2x.

Answer:

Given problem is
|1−i|x=2x
Now,
(12+(−1)2)x=2x
(1+1)x=2x
(2)x=2x
2x2=2x
x2=x
x2=0
x = 0 is the only possible solution to the given problem

Therefore, there are 0 number of non-zero integral solutions of the equation |1−i|x=2x

Question 12: 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:

It is given that
(a+ib)(c+id)(e+if)(g+ih)=A+iB,
Now, take mod on both sides
|(a+ib)(c+id)(e+if)(g+ih)|=|A+iB|
|(a+ib)||(c+id)||(e+if)||(g+ih)|=|A+iB| (∵|z1z2|=|z1||z2|)
(a2+b2)(c2+d2)(e2+f2)(g2+h2)=(A2+B2)
Square both sides. We will get

(a2+b2)(c2+d2)(e2+f2)(g2+h2)=(A2+B2)

Hence proved

Question 13: If (1+i1−i)m=1, then find the least positive integral value of m .

Answer:

Let
z=(1+i1−i)m
Now, multiply both numerator and denominator by (1+i)
We will get,
z=(1+i1−i×1+i1+i)m
=((1+i)212−i2)m
=(12+i2+2i1+1)m
=(1−1+2i2)m (∵i2=−1)
=(2i2)m
=im
We know that i4=1
Therefore, the least positive integral value of m is 4

Question 14: Let z1=2−i,z2=−2+i. Find

1.Re(z1z2z1)

2.Im(1z1z1―)

Answer(1):

It is given that

z 1 = 2 − i and z 2 = − 2 + i

Now,

z 1 z 2 = ( 2 − i ) ( − 2 + i ) = − 4 + 2 i + 2 i − i 2 = − 4 + 4 i + 1 = − 3 + 4 i

And

z ¯ 1 = 2 + i

Now,

z 1 z 2 z 1 = − 3 + 4 i 2 + i = − 3 + 4 i 2 + i × 2 − i 2 − i = − 6 + 3 i + 8 i − 4 i 2 2 2 − i 2 = − 6 + 11 i + 4 4 + 1 = − 2 + 11 i 5 = − 2 5 + i 11 5

Now,

Re ( z 1 z 2 z 1 ) = − 2 5

Therefore, the answer is

− 2 5

Answer(2):
It is given that

z 1 = 2 − i

Therefore,

z ¯ 1 = 2 + i

Now,

z 1 z ¯ 1 = ( 2 − i ) ( 2 + i ) = 2 2 − i 2 = 4 + 1 = 5 ( u sing ( a − b ) ( a + b ) = a 2 − b 2 )

Now,

1 z 1 z ¯ 1 = 1 5

Therefore,

Im ( 1 z 1 z ¯ 1 ) = 0

Therefore, the answer is 0