Chapter 5 – Complex Numbers and Quadratic Equations Questions and Answers: NCERT Solutions for Class 11 Maths

Exercise 5.1

1. Express the given complex number in the form a+ib:(5i)(−35i)a+ib:(5i)(−35i)
And evaluate

Ans:
Evaluate the complex number
(5i)(−35i)=−5×35×i×i(5i)(−35i)=−5×35×i×i
(5i)(−35i)=−3i2⋯[i2=−1](5i)(−35i)=−3i2⋯[i2=−1]
(5i)(−35i)=3(5i)(−35i)=3

2. Express the given complex number in the form a+ib:i9+i19a+ib:i9+i19
And evaluate

Ans:
Evaluate the complex number
i9+i19=i4×2+1+i4×4+3i9+i19=i4×2+1+i4×4+3
i9+i19=(i4)2.i+(i4)4.i3⋯[i4=1,i3=−1]i9+i19=(i4)2.i+(i4)4.i3⋯[i4=1,i3=−1]
i9+i19=0i9+i19=0

3. Express the given complex number in the form a+ib:i−39a+ib:i−39
And evaluate

Ans:
Evaluate the complex number
i−39=i4×9−3i−39=i4×9−3
i−39=(i4)−9.i−3i−39=(i4)−9.i−3
i−39=i⋯[i=−1]i−39=i⋯[i=−1]
i−39=ii−39=i

4. Express the given complex number in the form a+ib:3(7+i7)+i(7+i7)a+ib:3(7+i7)+i(7+i7)
And evaluate

Ans:
Evaluate the complex number
3(7+i7)+i(7+i7)=21+21i+7i+7i23(7+i7)+i(7+i7)=21+21i+7i+7i2
3(7+i7)+i(7+i7)=21+28i+7i2⋯[i2=−1]3(7+i7)+i(7+i7)=21+28i+7i2⋯[i2=−1]
3(7+i7)+i(7+i7)=14+28i3(7+i7)+i(7+i7)=14+28i

5. Express the given complex number in the form
a+ib:(1−i)−(−1+6i)a+ib:(1−i)−(−1+6i)
And evaluate

Ans:
Evaluate the complex number
(1−i)−(−1+6i)=1−i+1−i6(1−i)−(−1+6i)=1−i+1−i6
(1−i)−(−1+6i)=2−7i(1−i)−(−1+6i)=2−7i

6. Express the given complex number in the form a+ib:(15+i25)−(4+i52)a+ib:(15+i25)−(4+i52)
And evaluate

Ans:
Evaluate the complex number (15+i25)−(4+i52)=15+i25−4−i52(15+i25)−(4+i52)=15+i25−4−i52
(15+i25)−(4+i52)=−195+i[−2110](15+i25)−(4+i52)=−195+i[−2110]
(15+i25)−(4+i52)=−195−2110i(15+i25)−(4+i52)=−195−2110i

7. Express the given complex number in the form a+ib:[(13+i73)+(4+i13)−(−43+i)]a+ib:[(13+i73)+(4+i13)−(−43+i)]
And evaluate

Ans:
Evaluate the complex number
[(13+i73)+(4+i13)−(−43+i)]=13+i73+4+i13+43−i[(13+i73)+(4+i13)−(−43+i)]=13+i73+4+i13+43−i
[(13+i73)+(4+i13)−(−43+i)]=(13+4+43)+i(73+13−1)[(13+i73)+(4+i13)−(−43+i)]=(13+4+43)+i(73+13−1)
[(13+i73)+(4+i13)−(−43+i)]=173+i53[(13+i73)+(4+i13)−(−43+i)]=173+i53

8. Express the given complex number in the form a+ib:(1−i)4a+ib:(1−i)4
And evaluate

Ans:
Evaluate the complex number
(1−i)4=[1+i2−2i]2(1−i)4=[1+i2−2i]2
(1−i)4=[1−1−2i]2(1−i)4=[1−1−2i]2
(1−i)4=(−2i)×(−2i)(1−i)4=(−2i)×(−2i)
(1−i)4=−4(1−i)4=−4

9. Express the given complex number in the form a+ib:(13+3i)3a+ib:(13+3i)3
And evaluate

Ans:
Evaluate the complex number
(13+3i)3=(13)3+(3i)3+333i(13+3i)(13+3i)3=(13)3+(3i)3+333i(13+3i)
(13+3i)3=127−(27i)+3i(13+3i)(13+3i)3=127−(27i)+3i(13+3i)
(13+3i)3=−24227−26i(13+3i)3=−24227−26i

10. Express the given complex number in the form a+ib:(−2−13i)3a+ib:(−2−13i)3
And evaluate

Ans:
Evaluate the complex number
(−2−13i)3=(−1)3(2+13i)3(−2−13i)3=(−1)3(2+13i)3
(−2−13i)3=−(23+(i3)3+6i3(2+i3))(−2−13i)3=−(23+(i3)3+6i3(2+i3))
(−2−13i)3=−223−10727i(−2−13i)3=−223−10727i

11. Find the multiplicative inverse of the complex number
4−3i4−3i
And evaluate

Ans:
Let z=4−3iz=4−3i
Then,
z¯¯¯=4+3i&|z¯¯¯|=42+(−3)2=16+9=25z¯=4+3i&|z¯|=42+(−3)2=16+9=25
Therefore, the multiplicative inverse of 4−3i4−3i is given by
z−1=z¯¯¯|z|2=4+3i25=425+325iz−1=z¯|z|2=4+3i25=425+325i
Here we got final answer

12. Find the multiplicative inverse of the complex number 5–√+3i5+3i
And evaluate

Ans:
Let z=5–√+3iz=5+3i
Then,
z¯=5–√−3iand|z|2=(5–√)2+32=5+9=14z¯=5−3iand|z|2=(5)2+32=5+9=14
Therefore, the multiplicative inverse of 5–√+3i5+3i is given by
z−1=z¯|z|2=5–√−3i14=5–√14−3i14z−1=z¯|z|2=5−3i14=514−3i14
Here we got final answer

13. Find the multiplicative inverse of the complex number
−i−i
And evaluate

Ans:
Let z=−iz=−i
Then,
z¯=iand|z|2=12=1z¯=iand|z|2=12=1
Therefore, the multiplicative inverse of −i−i is given by z−1=z¯|z|2=i1=iz−1=z¯|z|2=i1=i
Here we got final answer

14. Express the following expression in the form of a+iba+ib (3+i5–√)(3−i5–√)(3–√+i2–√)−(3–√−i2–√)(3+i5)(3−i5)(3+i2)−(3−i2)
Evaluate

Ans:
The following expression (3+i5–√)(3−i5–√)(3–√+2–√i)−(3–√−i2–√)=(3)2−(i5–√)23–√+2–√i−3–√+2–√i(3+i5)(3−i5)(3+2i)−(3−i2)=(3)2−(i5)23+2i−3+2i
(3+i5–√)(3−i5–√)(3–√+2–√i)−(3–√−i2–√)=9−5i222–√i(3+i5–√)(3−i5–√)(3–√+2–√i)−(3–√−i2–√)=9−5(−1)22–√i(3+i5–√)(3−i5–√)(3–√+2–√i)−(3–√−i2–√)=−72i−−√2(3+i5)(3−i5)(3+2i)−(3−i2)=9−5i222i(3+i5)(3−i5)(3+2i)−(3−i2)=9−5(−1)22i(3+i5)(3−i5)(3+2i)−(3−i2)=−72i2
Here we got final answer

Exercise 5.2

1. Find the modulus and the argument of the complex number z=−1−i3–√z=−1−i3
Evaluate

Ans:
The complex number is
z=−1−i3–√z=−1−i3
Let rcosθ=-1andrsinθ=-3–√rcosθ=-1andrsinθ=-3
Squaring and adding
(rcosθ)2+(rsinθ)2=(−1)2+(−3–√)2(rcosθ)2+(rsinθ)2=(−1)2+(−3)2
r2(cos2θ+sin2θ)=1+3r2=4[cos2θ+sin2θ=1]r2(cos2θ+sin2θ)=1+3r2=4[cos2θ+sin2θ=1]
r=4–√=2[Conventionally,r0]r=4=2[Conventionally,r0]
Modulus=22cosθ=-1and2sinθ=-3–√cosθ=−12andsinθ=−3–√2Modulus=22cosθ=-1and2sinθ=-3cosθ=−12andsinθ=−32
Since both the values of sinθandcosθsinθandcosθ negative and sinθandcosθsinθandcosθ are negative in 3rd quadrant,
Argument=-(π-π3)=−2π3Argument=-(π-π3)=−2π3
Thus, the modulus and argument of the complex number −1−3–√iare2and-2π3−1−3iare2and-2π3
Respectively

2. Find the modulus and the argument of the complex number
z=−3–√+iz=−3+i
Evaluate

Ans:
The complex number is
z=−3–√+iz=−3+i
Let rcosθ=-3–√andrsinθ=1rcosθ=-3andrsinθ=1
squaring and adding
(rcosθ)2+(rsinθ)2=(−3–√)2+(−1)2(rcosθ)2+(rsinθ)2=(−3)2+(−1)2
r2=3+1=4LLL[cos2θ+sin2θ=1]r=4–√=2LLL[Conventionally,r0]r2=3+1=4LLL[cos2θ+sin2θ=1]r=4=2LLL[Conventionally,r0]
Modulus=22cosθ=-3–√and2sinθ=1Modulus=22cosθ=-3and2sinθ=1
cosθ=−3–√2andsinθ=12θ=π-π6=5π6LL[Asθlies in the II quadrant]cosθ=−32andsinθ=12θ=π-π6=5π6LL[Asθlies in the II quadrant]
Thus, the modulus and argument of the complex number −3–√+iare2and5π6−3+iare2and5π6
Respectively

3. Convert the given complex number in polar form
1−i1−i
And evaluate

Ans:
The complex number is
1−i1−i
Let rcosθ=1andrsinθ=−1rcos⁡θ=1andrsin⁡θ=−1
squaring and adding
r2cos2θ+r2sin2θ=12+(−1)2⇒r2(cos2θ+sin2θ)=1+1r2cos2θ+r2sin2θ=12+(−1)2⇒r2(cos2θ+sin2θ)=1+1
r2=2r=2–√[Conventionally,r>0]r2=2r=2[Conventionally,r>0]
2–√cosθ=1and2–√sinθ=-1cosθ=12–√andsinθ=-12–√θ=-π4[Asθliesin the IV quadrant]2cosθ=1and2sinθ=-1cosθ=12andsinθ=-12θ=-π4[Asθliesin the IV quadrant]
1-i=rcosθ+irsinθ=2–√cos(−π4)+i2–√sin(−π4)=2–√[cos(−π4)+isin(−π4)]1-i=rcosθ+irsinθ=2cos(−π4)+i2sin(−π4)=2[cos(−π4)+isin(−π4)]
Required polar form

4. Convert the given complex number in polar form
−1+i−1+i
And evaluate

Ans:
The complex number is
−1+i−1+i
Let rcosθ=-1andrsinθ=1rcosθ=-1andrsinθ=1
Squaring and adding
r2cos2θ+r2sin2θ=(-1)2+12r2(cos2θ+sin2θ)=1+1r2=2r=2–√r2cos2θ+r2sin2θ=(-1)2+12r2(cos2θ+sin2θ)=1+1r2=2r=2
2–√cosθ=-1and2–√sinθ=12cosθ=-1and2sinθ=1
cosθ=-12–√and2–√sinθ=1θ=π-π4=3π4L[Asθlies in the II quadrant]cosθ=-12and2sinθ=1θ=π-π4=3π4L[Asθlies in the II quadrant]
It can be written,
-1+i=rcosθ+irsinθ=2–√cos3π4+i2–√sin3π4=2–√(cos3π4+isin3π4)-1+i=rcosθ+irsinθ=2cos3π4+i2sin3π4=2(cos3π4+isin3π4)
Required polar form

5. Convert the given complex number in polar form −1−i−1−i
And evaluate

Ans:
The complex number is
−1−i−1−i
Let rcosθ=-1andrsinθ=-1rcosθ=-1andrsinθ=-1
Squaring and adding
r2cos2θ+r2sin2θ=(-1)2+(−1)2r2(cos2θ+sin2θ)=1+1r2=2r=2–√r2cos2θ+r2sin2θ=(-1)2+(−1)2r2(cos2θ+sin2θ)=1+1r2=2r=2
2–√cosθ=-1and2–√sinθ=-1cosθ=-12–√andsinθ=-12–√θ–(π-π4)−−3π4[As0lies in the III quadrant]2cosθ=-1and2sinθ=-1cosθ=-12andsinθ=-12θ–(π-π4)−−3π4[As0lies in the III quadrant]
-1-i=rcosθ+irsinθ=2–√cos−3π4+i2–√sin−3π4=2–√(cos−3π4+isin−3π4)-1-i=rcosθ+irsinθ=2cos−3π4+i2sin−3π4=2(cos−3π4+isin−3π4)
Required polar form

6. Convert the given complex number in polar form
−3−3
And evaluate

Ans:
The complex number is
−3−3
Let rcosθ=-3andrsinθ=0rcosθ=-3andrsinθ=0
Squaring and adding
r2cos2θ+r2sin2θ=(-3)2r2(cos2θ+sin2θ)=9r2cos2θ+r2sin2θ=(-3)2r2(cos2θ+sin2θ)=9
r2=9r=9–√=3r2=9r=9=3
3cosθ=-3and3sinθ=0cosθ=-1andsin=0θ=π3cosθ=-3and3sinθ=0cosθ=-1andsin=0θ=π
-3=rcosθ+irsinθ=3cosπ+i3sinπ=3(cosπ+isinπ)-3=rcosθ+irsinθ=3cosπ+i3sinπ=3(cosπ+isinπ)
Required polar form

7. Convert the given complex number in polar form 3–√+i3+i
And evaluate

Ans:
The complex number is
3–√+i3+i
Let rcosθ=3–√andrsinθ=1rcosθ=3andrsinθ=1
Squaring and adding
r2cos2θ+r2sin2θ=(3–√)2+12r2(cos2θ+sin2θ)=3+1r2=4r=4–√=2r2cos2θ+r2sin2θ=(3)2+12r2(cos2θ+sin2θ)=3+1r2=4r=4=2
2cosθ=3–√and2sinθ=1cosθ=3–√2andsinθ=12θ=π6[Asθlies in the I quadrant]2cosθ=3and2sinθ=1cosθ=32andsinθ=12θ=π6[Asθlies in the I quadrant]
3–√+i=rcosθ+irsinθ=2cosπ6+i2sinπ6=2(cosπ6+isinπ6)3+i=rcosθ+irsinθ=2cosπ6+i2sinπ6=2(cosπ6+isinπ6)
Required polar form

8. Convert the given complex number in polar form
ii
And evaluate

Ans:
The complex number is
ii
Let rcosθ=0andrsinθ=1rcosθ=0andrsinθ=1
Squaring and adding
r2cos2θ+r2sin2θ=02+12r2(cos2θ+sin2θ)=1r2cos2θ+r2sin2θ=02+12r2(cos2θ+sin2θ)=1
r2=1r=1–√=1[Conventionally,r0]r2=1r=1=1[Conventionally,r0]
cosθ=0andsinθ=1θ=π2i=rcosθ+irsinθ=cosπ2+isinπ2cosθ=0andsinθ=1θ=π2i=rcosθ+irsinθ=cosπ2+isinπ2
Required polar form

Exercise 5.3

1. Solve the equation
x2+3=0x2+3=0
And evaluate

Ans:
Quadratic equation x2+3=0x2+3=0
General form ax2+bx+c=0ax2+bx+c=0
We obtain a=1,b=0,andc=3a=1,b=0,andc=3
Therefore, the discriminant of the given equation is
D=b2−4ac=02−4×1×3=-12D=b2−4ac=02−4×1×3=-12
Therefore, the required solutions are
=−b±D−−√2a=±−12−−−−√2×1=±12−−√i2=±23–√i2=±3–√i=−b±D2a=±−122×1=±12i2=±23i2=±3i

2. Solve the equation
2×2+x+1=02×2+x+1=0
And evaluate

Ans:
Quadratic equation 2×2+x+1=02×2+x+1=0
General form ax2+bx+c=0ax2+bx+c=0
We obtain a=2,b=1,andc=1a=2,b=1,andc=1
Therefore, the discriminant of the given equation is
D=b2−4ac=12−4×2×1=-7D=b2−4ac=12−4×2×1=-7
Therefore, the required solutions are
=−b±D−−√2a=±−7−−−√2×2=±7–√i4=−b±D2a=±−72×2=±7i4

3. Solve the equation
x2+3x+9=0x2+3x+9=0
And evaluate

Ans:
Quadratic equation x2+3x+9=0x2+3x+9=0
General form ax2+bx+c=0ax2+bx+c=0
We obtain a=1,b=3,andc=9a=1,b=3,andc=9
Therefore, the discriminant of the given equation is
D=b2−4ac=32−4×1×9=-27D=b2−4ac=32−4×1×9=-27
Therefore, the required solutions are
=−b±D−−√2a=−3±−27−−−−√2×1=−3±33–√i2=−b±D2a=−3±−272×1=−3±33i2

4. Solve the equation
−x2+x−2=0−x2+x−2=0
And evaluate

Ans:
Quadratic equation −x2+x−2=0−x2+x−2=0
General form ax2+bx+c=0ax2+bx+c=0
We obtain a=−1,b=1,andc=-2a=−1,b=1,andc=-2
Therefore, the discriminant of the given equation is
D=b2−4ac=12−4×-1×-2=-7D=b2−4ac=12−4×-1×-2=-7
Therefore, the required solutions are
=−b±D−−√2a=−1±−7−−−√2×-1=−1±7–√i−2=−b±D2a=−1±−72×-1=−1±7i−2

5. Solve the equation
x2+3x+5=0x2+3x+5=0
And evaluate

Ans:
Quadratic equation x2+3x+5=0x2+3x+5=0
General form ax2+bx+c=0ax2+bx+c=0
We obtain a=1,b=3,andc=5a=1,b=3,andc=5
Therefore, the discriminant of the given equation is
D=b2−4ac=32−4×1×5=-11D=b2−4ac=32−4×1×5=-11
Therefore, the required solutions are
=−b±D−−√2a=−3±−11−−−−√2×1=−3±11−−√i2=−b±D2a=−3±−112×1=−3±11i2

6. Solve the equation
x2−x+2=0x2−x+2=0
And evaluate

Ans:
Quadratic equation x2−x+2=0x2−x+2=0
General form ax2+bx+c=0ax2+bx+c=0
We obtain a=1,b=3−1,andc=2a=1,b=3−1,andc=2
Therefore, the discriminant of the given equation is
D=b2−4ac=(−1)2−4×1×2=-7D=b2−4ac=(−1)2−4×1×2=-7
Therefore, the required solutions are
=−b±D−−√2a=−(−1)±−7−−−√2×1=1±7–√i2=−b±D2a=−(−1)±−72×1=1±7i2

7. Solve the equation
2–√x2+x+2–√=02×2+x+2=0
And evaluate

Ans:
Quadratic equation 2–√x2+x+2–√=02×2+x+2=0
General form ax2+bx+c=0ax2+bx+c=0
We obtain a=2–√,b=1,andc=2–√a=2,b=1,andc=2
Therefore, the discriminant of the given equation is
D=b2−4ac=12−4×2–√×2–√=−7D=b2−4ac=12−4×2×2=−7
Therefore, the required solutions are
=−b±D−−√2a=−1±−7−−−√2×2–√=−1±7–√i22–√=−b±D2a=−1±−72×2=−1±7i22

8. Solve the equation
3–√x2−2–√x+33–√=03×2−2x+33=0
And evaluate

Ans:
Quadratic equation 3–√x2−2–√x+33–√=03×2−2x+33=0
General form ax2+bx+c=0ax2+bx+c=0
We obtain a=3–√,b=−2–√,andc=33–√a=3,b=−2,andc=33
D=b2−4ac=(−2–√)2−4×3–√×33–√=−34D=b2−4ac=(−2)2−4×3×33=−34 D=b2−4ac=(−2–√)2−4×3–√×33–√=−34D=b2−4ac=(−2)2−4×3×33=−34
Therefore, the required solutions are
=−b±D−−√2a=−(−2–√)±−34−−−−√2×3–√=2–√±34−−√i23–√=−b±D2a=−(−2)±−342×3=2±34i23

9. Solve the equation
x2+x+12–√=0x2+x+12=0
And evaluate

Ans:
Quadratic equation x2+x+12–√=0x2+x+12=0
General form ax2+bx+c=0ax2+bx+c=0
We obtain a=2–√,b=2–√,andc=1a=2,b=2,andc=1
Therefore, the discriminant of the given equation is
D=b2−4ac=(2–√)2−4×2–√×1=2-42–√D=b2−4ac=(2)2−4×2×1=2-42
Therefore, the required solutions are
=−b±D−−√2a=−(2–√)±2−42–√−−−−−−−√2×2–√=−1±(22–√−1−−−−−−√)i2=−b±D2a=−(2)±2−422×2=−1±(22−1)i2

10. Solve the equation
x2+x2–√+1=0x2+x2+1=0
And evaluate

Ans:
Quadratic equation x2+x2–√+1=0x2+x2+1=0
General form ax2+bx+c=0ax2+bx+c=0
We obtain a=2–√,b=1,andc=2–√a=2,b=1,andc=2
Therefore, the discriminant of the given equation is
D=b2−4ac=(1)2−4×2–√×2–√=−7D=b2−4ac=(1)2−4×2×2=−7
Therefore, the required solutions are
=−b±D−−√2a=−(1)±−7−−−√2×2–√=−1±7–√i22–√=−b±D2a=−(1)±−72×2=−1±7i22