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Learn/ 11th Class Mathematics/ Unit 1 /Numericals

Unit 1: Complex Numbers — Exercise 1 3

11th Class Mathematics · Unit 1: Complex Numbers

1.(i) Factorize: a^2 + 4b^2 (ii) Factorize: 9a^2 + 16b^2 (iii) Factorize: 3x^2 + 3y^2 (iv) Factorize: 144x^2 + 225y^2 (v) Factorize: z^2 - 2iz - 1 (vi) Factorize: z^2 + 6z + 13 (vii) Factorize: z^2 + 4z + 5 (viii) Factorize: 2z^2 - 22z + 65
(i) a2+4b2 = a2-4b2i2
= (a)2-(2bi)2
Result
= (a-2bi)(a+2bi)
(ii) 9a2+16b2 = 9a2-16b2i2
= (3a)2-(4bi)2
= (3a-4bi)(3a+4bi)
(iii) 3x2+3y2 = 3(x2+y2)
= 3(x2-i2y2) = 3left[(x)2-(iy)2right]
= 3(x+iy)(x-iy)
(iv) 144x2+225y2 = 144x2-225y2i2 = (12x)2-(15iy)2
= (12x-15iy)(12x+15iy)
= 9(4x-5iy)(4x+5iy)
(v) z2-2iz-1 = z2-2iz+i2 = z2-iz-iz+i2
= z(z-i)-i(z-i)
= (z-i)(z-i)
(vi) z2+6z+13 = z2+6z+9+4 = (z+3)2-4i2
= (z+3)2-(2i)2
= (z+3+2i)(z+3-2i)
(vii) z2+4z+5 = z2+4z+4+1 = (z+2)2-i2
= (z+2+i)(z+2-i)
Formula
(viii) Quadratic formula z = frac{22 ± sqrt{(-22)2-4(2)(65)}{2(2)}
z = frac{22 ± sqrt{484-520}{4} = 22 ± 6i4 = 11 ± 3i2
Result
Factors left(z-11+3i2right)left(z-11-3i2right)
2.Factorize the following polynomial into its linear factors: (i) z^3 + 8 (ii) z^3 + 27 (iii) z^3 - 2z^2 + 16z - 32 (iv) z^4 + 21z^2 - 100 (v) z^4 - 16 (vi) z^4 + 3z^2 - 4 (vii) z^4 + 5z^2 + 6 (viii) z^4 - 32z^2 - 3969
Formula
(i) Sum of cubes z3+8 = (z)3+(2)3 = (z+2)(z2-2z+4)
Quadratic formula for z^2-2z+4 z = frac{-(-2) ± sqrt{(-2)2-4(1)(4)}{2(1)} = frac{2 ± sqrt{4-16}{2}
z = 2 ± 2sqrt3 i2 = 1 ± sqrt3 i
Result
(z+2)left(z-1-sqrt3 iright)left(z-1+sqrt3 iright)
Formula
(ii) Sum of cubes z3+27 = (z)3+(3)3 = (z+3)(z2-3z+9)
Quadratic formula for z^2-3z+9 z = frac{-(-3) ± sqrt{(-3)2-4(1)(9)}{2(1)} = frac{3 ± sqrt{9-36}{2}
z = frac{3 ± sqrt{27}i}{2} = 3 ± 3sqrt3 i2
Result
(z+3)left(z-3+3sqrt3 i2right)left(z-3-3sqrt3 i2right)
(iii) z3-2z2+16z-32 = z2(z-2)+16(z-2)
= (z-2)(z2+16) = (z-2)left[(z)2-(4i)2right]
= (z-2)(z-4i)(z+4i)
(iv) z4+21z2-100 = z4+25z2-4z2-100
= z2(z2+25)-4(z2+25) = (z2+25)(z2-4)
= (z2-22)(z2-25i2) = (z-2)(z+2)(z-5i)(z+5i)
(v) z4-16 = (z2)2-(4)2 = (z2-4)(z2+4)
= left[(z)2-(2)2right]left[(z)2-(2i)2right]
= (z+2)(z-2)(z+2i)(z-2i)
(vi) z4+3z2-4 = z4+4z2-z2-4
= z2(z2+4)-1(z2+4) = (z2-1)(z2+4)
= (z-1)(z+1)(z+2i)(z-2i)
(vii) z4+5z2+6 = z4+3z2+2z2+6
= z2(z2+3)+2(z2+3) = (z2+2)(z2+3)
= (z+sqrt2 i)(z-sqrt2 i)(z+sqrt3 i)(z-sqrt3 i)
(viii) z4-32z2-3969 = z4-81z2+49z2-3969
= z2(z2-81)+49(z2-81) = (z2-81)(z2+49)
= (z2-92)(z2-(7i)2) = (z-9)(z+9)(z-7i)(z+7i)
3.Find the roots of z^4 + 7z^2 - 144 = 0 and hence express it as a product of linear factors.
z4+7z2-144=0 ⇒ z4+16z2-9z2-144=0
z2(z2+16)-9(z2+16)=0 ⇒ (z2-9)(z2+16)=0
left[(z)2-(3)2right]left[(z)2-(4i)2right]=0 ⇒ (z+3)(z-3)(z+4i)(z-4i)=0
Result
Roots z = 3,-3,4i,-4i
4.Solve the following complex quadratic equation by completing square method: (i) 2z^2 - 3z + 4 = 0 (ii) z^2 - 6z + 30 = 0 (iii) 3z^2 - 18z + 50 = 0 (iv) z^2 + 4z + 13 = 0 (v) 2z^2 + 6z + 9 = 0 (vi) 3z^2 - 5z + 7 = 0
(i) 2z2-3z=-4 ⇒ z2-32z=-2
Complete the square z2-frac32 z+916=-2+916 ⇒ left(z-frac34right)2=-2316
Result
z-frac34= ± sqrt{-2316 ⇒ z=frac{3 ± sqrt{23}i}{4}
(ii) z2-6z=-30 ⇒ z2-6z+9=-30+9
(z-3)2=-21 ⇒ z-3= ± sqrt{-21}
z = 3 ± sqrt{21}i
(iii) 3z2-18z=-50 ⇒ z2-6z=-503
Complete the square z2-6z+9=-503+9 ⇒ (z-3)2=-233
z-3= ± sqrt{233i ⇒ z=3 ± sqrt{233i
(iv) z2+4z=-13 ⇒ z2+4z+4=-13+4
(z+2)2=-9 ⇒ z+2= ± 3i
z=-2 ± 3i
(v) 2z2+6z=-9 ⇒ z2+3z=-frac92
Complete the square z2+3z+frac94=-frac92+frac94 ⇒ left(z+frac32right)2=-frac94
z+frac32= ± frac32 i ⇒ z=-frac32 ± frac32 i
(vi) 3z2-5z=-7 ⇒ z2-frac53 z=-frac73
Complete the square z2-frac53 z+2536=-frac73+2536 ⇒ left(z-frac56right)2=-5936
z-frac56= ± frac{sqrt{59}{6}i ⇒ z=frac56 ± frac{sqrt{59}{6}i
5.Solve the following equation: (i) 2z^4 - 32 = 0 (ii) 3z^5 - 243z = 0 (iii) 5z^5 - 5z = 0 (iv) z^3 - 5z^2 + z - 5 = 0 (v) 4z^4 - 25z^2 - 21 = 0 (vi) z^3 + z^2 + z + 1 = 0
(i) 2(z4-16)=0 ⇒ (z2)2-(4)2=0 ⇒ (z2-4)(z2+4)=0
left[(z)2-(2)2right]left[(z)2-(2i)2right]=0 ⇒ (z-2)(z+2)(z-2i)(z+2i)=0
Result
z=2,-2,2i,-2i
(ii) 3z(z4-81)=0 ⇒ zleft[(z2)2-(9)2right]=0
z(z2-9)(z2+9)=0 ⇒ zleft[(z)2-(3)2right]left[(z)2-(3i)2right]=0
z(z+3)(z-3)(z+3i)(z-3i)=0 ⇒ z=0,-3,3,3i,-3i
(iii) 5z(z4-1)=0 ⇒ zleft[(z2)2-(1)2right]=0
z(z2-1)(z2+1)=0 ⇒ z(z-1)(z+1)(z-i)(z+i)=0
z=0, ± 1, ± i
(iv) z2(z-5)+1(z-5)=0 ⇒ (z-5)(z2+1)=0
(z-5)(z-i)(z+i)=0
z=5, ± i
Working
(v) Let z^2 = t 4t2-25t-21=0
Formula
Quadratic formula t=frac{-(-25) ± sqrt{(-25)2-4(4)(-21)}{2(4)} = frac{25 ± sqrt{625+336}{8}=25 ± 318
t=7 or t=-frac34
Result
z2=7 ⇒ z= ± sqrt7;quad z2=-frac34 ⇒ z= ± sqrt32i
(vi) z2(z+1)+1(z+1)=0 ⇒ (z+1)(z2+1)=0
(z+1)(z-i2) i.e. (z+1)(z-i)(z+i)=0
Roots z=-1,i,-i
6.Find a polynomial P(z) of degree 3 with zeros 3, -2i, 2i and satisfying P(1) = 20.
Formula
P(z) = a(z-3)(z+2i)(z-2i) = a(z-3)(z2-4i2) = a(z-3)(z2+4)
P(z) = a(z3+4z-3z2-12) = a(z3-3z2+4z-12) quad (i)
Working
Using P(1)=20 20 = a[1-3+4-12] = a(-10) ⇒ a=-2
Result
P(z) = -2(z3-3z2+4z-12) = -2z3+6z2-8z+24
7.Find a polynomial P(z) of degree 4 with zeros 2i, -2i, 1, -1, and satisfying P(2) = 240.
Formula
P(z) = a(z+1)(z-1)(z+2i)(z-2i) = a(z2-1)(z2-4i2)
= a(z2-1)(z2+4) = a(z4+4z2-z2-4) = a(z4+3z2-4) quad (i)
Working
Using P(2)=240 240 = aleft[(2)4+3(2)2-4right] = a[16+12-4] = a(24) ⇒ a=10
Result
P(z) = 10(z4+3z2-4) = 10z4+30z2-40
8.Find a polynomial P(z) of degree 4 with zeros 4, -4, 1+i, 1-i, and satisfying P(2) = 72.
Formula
P(z) = a(z-4)(z+4)big(z-(1+i)big)big(z-(1-i)big) = a(z2-16)left[(z-1)2-i2right]
= a(z2-16)(z2-2z+2) = a(z4-2z3+2z2-16z2+32z-32)
= a(z4-2z3-14z2+32z-32) quad (i)
Working
Using P(2)=72 72 = aleft[24-2(2)3-14(2)2+32(2)-32right] = a[-24] ⇒ a=-3
Result
P(z) = -3(z4-2z3-14z2+32z-32) = -3z4+6z3+42z2-96z+96