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 2x2+x+1=02x2+x+1=0 And evaluate

Ans: Quadratic equation 2x2+x+1=02x2+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–√=02x2+x+2=0 And evaluate

Ans: Quadratic equation 2–√x2+x+2–√=02x2+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–√=03x2−2x+33=0 And evaluate

Ans: Quadratic equation 3–√x2−2–√x+33–√=03x2−2x+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