11th Class Mathematics Exercise 1.2

1.Find the values of x and y: (i) x + iy + 2 - 3i = i(5-i)(3+4i) (ii) (x+iy)(1-i) = (2-3i)(-5+5i)(-i\frac{3}{5}) (iii) \frac{x}{2+i}+\frac{y}{3-i}=4+5i

Given

(i) Equation x+iy+2-3i = i(5-i)(3+4i)

(x+2)+i(y-3) = ileft[15+20i-3i-4i²right]

(x+2)+i(y-3) = ileft[15+4+17iright]

(x+2)+i(y-3) = -17+19i

Comparing real and imaginary parts x+2=-17 ;quad y-3=19

Result

x=-19 ;quad y=22

Given

(ii) Equation (x+iy)(1-i) = (2-3i)(-5+5i)left(-i35right)

=(2-3i) · 5(-1+i)left(-i35right)

x-xi+yi-i²y=(2-3i)(3i-3i²)

x-xi+yi+y=(2-3i)(3i+3)

(x+y)+i(y-x)=6i+6-9i²-9i

(x+y)+i(y-x)=15-3i

Comparing real and imaginary parts x+y=15 (i) ;quad -x+y=-3 (ii)

Adding (i) and (ii) 2y=12 ⇒ y=6

Put in (i) x+6=15 ⇒ x=9

Result

x=9, y=6

Given

(iii) Equation x2+i+y3-i=4+5i

x(3-i)+y(2+i)(2+i)(3-i)=4+5i

3x-xi+2y+yi=(4+5i)(6+1+i)

(3x+2y)+(y-x)i=28+4i+35i+5i²

(3x+2y)+i(y-x)=23+39i

Comparing real and imaginary parts 3x+2y=23 (i) ;quad -x+y=39 (ii)

From (ii) y=39+x

Substitute in (i) 3x+78+2x=23 ⇒ 5x=-55 ⇒ x=-11

Put in (ii) -(-11)+y=39 ⇒ y=28

Result

x=-11, y=28

2.If z1 = -13 + 24i and z2 = x + yi, find the values of x and y such that z1 - z2 = -27 + 15i

Given

Given z₁-z₂=-27+15i

Working

Replacing z1 and z2 (-13+24i)-(x+iy)=-27+15i

Given

Comparing real and imaginary parts -13-x=-27 ;quad 24-y=15

Result

x=14 ;quad y=9

3.Find the value of x and y if: (i) (x+iy)^2 = 25 + 60i (ii) (x+iy)^2 = 64 + 48i (iii) (x+iy)^2 = \frac{2i-3}{3+i}

Given

(i) Given (x+iy)²=25+60i

x²-y²+2xyi=25+60i

Comparing real and imaginary parts x²-y²=25 (i) ;quad 2xy=60 (ii)

From (ii) y=30x

Substitute in (i) x²-900x²=25 ⇒ x⁴-25x²-900=0

x²(x²-45)+20(x²-45)=0 ⇒ x²=45 or x²=-20 (not possible)

Put x^2=45 in (i) 45-y²=25 ⇒ y²=20 ⇒ y=2sqrt5

Result

x=3sqrt5, y=2sqrt5 or x=-3sqrt5, y=-2sqrt5

Given

(ii) Given (x+iy)²=64+48i

x²-y²+2xyi=64+48i

Comparing real and imaginary parts x²-y²=64 (i) ;quad 2xy=48 (ii)

From (ii) y=24x

Substitute in (i) x²-576x²=64 ⇒ x⁴-64x²-576=0

x²(x²-72)+8(x²-72)=0 ⇒ x²=72 or x²=-8 (not possible)

Put x^2=72 in (i) 72-y²=64 ⇒ y²=8 ⇒ y=2sqrt2

Result

x=6sqrt2, y=2sqrt2 or x=-6sqrt2, y=-2sqrt2

Given

(iii) Given (x+iy)²=2i-33+i

x²-y²+2ixy=2i-33+i × 3-i3-i=6i-2i²-9+3i9-i²

x²-y²+2ixy=9i+2-910=-710+910i

Comparing real and imaginary parts x²-y²=-710 (i) ;quad 2xy=910 ⇒ xy=920 (ii)

From (ii) y=920x

Substitute in (i) x²-81400x²=-710 ⇒ 400x⁴+280x²-81=0

Quadratic formula for x^2 x²=frac{-280 ± sqrt{280²-4(400)(-81)}{2(400)}=frac{-280 ± sqrt{208000}{800}=frac{-7 ± sqrt{130}{20}

Take positive root x²=frac{-7+sqrt{130}{20} ≈ 0.22 ⇒ x ≈ ± 0.46

y=920x ⇒ y ≈ ± 0.97

Result

x ≈ 0.46, y ≈ 0.97 or x ≈ -0.46, y ≈ -0.97

4.If z1 = 2 + 3i and z2 = 1 - alpha, find the value of alpha such that Im(z1.z2) = 7

Given

Given Im(z₁ z₂)=7

Working

Imleft[(2+3i)(1-α)right]=7

Imleft[2(1-α)+3i(1-α)right]=7 ⇒ 3(1-α)=7

1-α=73

Result

α=1-73=-43

5.If z1 = x + yi and z2 = a + bi, find x, y, a and b such that z1 + z2 = 10 + 4i and z1 - z2 = 6 + 2i.

Given

Given z₁+z₂=10+4i ;quad z₁-z₂=6+2i

(x+a)+(y+b)i=10+4i ⇒ x+a=10 (i) ;quad y+b=4 (ii)

(x-a)+(y-b)i=6+2i ⇒ x-a=6 (iii) ;quad y-b=2 (iv)

Adding (i) and (iii) 2x=16 ⇒ x=8 ⇒ a=2

Adding (ii) and (iv) 2y=6 ⇒ y=3 ⇒ b=1

Result

x=8, y=3, a=2, b=1

6.Show that for all z1, z2 in C, conjugate(z1 z2) = conjugate(z1) . conjugate(z2)

Given

Let z₁=a+bi,quad z₂=c+di

z₁ z₂=(a+bi)(c+di)=(ac-bd)+(ad+bc)i

L.H.S. overline{z₁ z₂}=(ac-bd)-(ad+bc)i

=ac-adi-bci+bdi²=a(c-di)-ib(c-di)

Result

=(a-ib)(c-di)=overline{z₁} · overline{z₂}=R.H.S.

7.Find the square root of the following complex numbers: (i) -7 - 24i (ii) 8 - 6i (iii) -15 - 36i (iv) 119 + 120i

Given

(i) Let z=x+iy=-7-24i ⇒ x=-7, y=-24<0

Modulus |z|=sqrt{49+576}=25

Formula

Square root formula sqrt{z}= ± left(sqrt{|z|+x2+iy|y|sqrt{|z|-x2right)

Working

sqrt{-7-24i}= ± left(sqrt{25-72+i(-24)24sqrt{25+72right)

= ± (3-4i)

Result

3-4i or -3+4i

Given

(ii) Let z=x+iy=8-6i ⇒ x=8, y=-6<0

Modulus |z|=sqrt{64+36}=10

Formula

Square root formula sqrt{z}= ± left(sqrt{|z|+x2+iy|y|sqrt{|z|-x2right)

Working

sqrt{8-6i}= ± left(sqrt{10+82+(-6)i6sqrt{10-82right)

= ± (3-i)

Result

3-i or -3+i

Given

(iii) Let z=x+iy=-15-36i ⇒ x=-15, y=-36<0

Modulus |z|=sqrt{225+1296}=39

Formula

Square root formula sqrt{z}= ± left(sqrt{|z|+x2+iy|y|sqrt{|z|-x2right)

Working

sqrt{-15-36i}= ± left(sqrt{39-152+(-36)i36sqrt{39+152right)

= ± (2sqrt3-3sqrt3 i)

Result

2sqrt3-3sqrt3 i or -2sqrt3+3sqrt3 i

Given

(iv) Let z=x+iy=119+120i ⇒ x=119, y=120>0

Modulus |z|=sqrt{119²+120²}=169

Formula

Square root formula sqrt{z}= ± left(sqrt{|z|+x2+iy|y|sqrt{|z|-x2right)

Working

sqrt{119+120i}= ± left(sqrt{169+1192+i(120)120sqrt{169-1192right)

= ± (12+5i)

Result

12+5i or -12-5i

8.Find the square root of 13 - 20\sqrt3 i and represent them on an Argand diagram.

Given

Let z=x+iy=13-20sqrt3 i ⇒ x=13, y=-20sqrt3<0

Modulus |z|=sqrt{13²+(20sqrt3)²}=37

Formula

Square root formula sqrt{z}= ± left(sqrt{|z|+x2+iy|y|sqrt{|z|-x2right)

Working

sqrt{13-20sqrt3 i}= ± left(sqrt{37+132+i(-20sqrt3)20sqrt3sqrt{37-132right)

= ± (5-2sqrt3 i)

Result

5-2sqrt3 i or -5+2sqrt3 i (plotted on Argand diagram)

9.Find the value of x and y if: (-7+i)(x+iy)+(-1-5i) = i(11-i)

Given

Given (-7+i)(x+iy)+(-1-5i)=i(11-i)

-7x-7yi+xi+yi²-1-5i=11i-i²

(-7x-y-1)+i(-7y+x-5)=1+11i

Comparing real and imaginary parts -7x-y-1=1 (i) ;quad x-7y-5=11 (ii)

From (i) y=-7x-2

Substitute in (ii) x-7(-7x-2)=16 ⇒ 50x=2 ⇒ x=125

Put in (ii) 7y=x-16=125-16=-39925 ⇒ y=-5725

Result

x=125, y=-5725

10.Find the value of x and y if: (5-2i)(x+iy)+3 = i(11-i)-4i

Given

Given (5-2i)(x+iy)+3=i(11-i)-4i

5x+5yi-2xi-2yi²+3=11i-i²-4i

(5x+2y+3)+i(5y-2x)=7i+1

Comparing real and imaginary parts 5x+2y+3=1 (i) ;quad 5y-2x=7 (ii)

Multiply (i) by 2, (ii) by 5 and add 10x+4y=-4 ;quad -10x+25y=35 ⇒ 29y=31 ⇒ y=3129

Put in (i) 5x+2left(3129right)=-2 ⇒ 5x=-12029 ⇒ x=-2429

Result

x=-2429, y=3129

11.Find the real values of u and v if: \frac{u-2}{2+i}+\frac{v-3}{2-i}=4i

Given

Given u-22+i+v-32-i=4i

(u-2)(2-i)+(v-3)(2+i)(2+i)(2-i)=4i

2u-iu-4+2i+2v+iv-6-3i=4i(4-i²)

2u+2v-10+i(v-u-1)=20i

Comparing real and imaginary parts 2u+2v-10=0 ⇒ u+v-5=0 (i) ;quad -u+v-1=20 ⇒ -u+v-21=0 (ii)

Adding (i) and (ii) 2v-26=0 ⇒ v=13

Put in (i) u+13-5=0 ⇒ u=-8

Result

u=-8, v=13

12.If z1 = 4 + 5i and z2 = alpha - 2i, find the real values of alpha such that Re(z1 z2) = 20.

Given

Given Re(z₁ z₂)=20

Working

Releft[(4+5i)(α-2i)right]=20

Releft[4α-8i+5αi+10right]=20 ⇒ 4α+10=20

Result

α=104=52

Unit 1: Complex Numbers — Exercise 1 2