CLASS XI CHAPTER 5 COMPLEX NUMBERS AND QUADRATIC EQUATIONS RR ACADEMY MEERUT

Chapter 5: Complex Numbers and Quadratic Equations

Exercise: 5.1

Q.1: Express the following complex number in x + iy form; (4i)(−74i)

Sol:

(4i)(−74i)  = -4 × 74 × i × i = -7 i² = -7(-1)   [ Since, i² = -1 ]

= 7

Q.2: Express the following complex number in x + iy form; i19+i9 

Sol:

i19+i9=i4×2+1+i4×4+3 = (i4)2⋅i+(i4)4⋅i3

=1×i+1×(−i)   [ Since,  i⁴ = 1, i³ = -1 ]

= i + (-i)

= 0

Q.3: Express the following complex number in x + iy form; i−39

Sol:

i−39=i−4×9–3 =(i4)−9⋅i−3

=(1)−9⋅i−3   [ Since, i⁴ = 1 ]   

=1i3=1−i   [ Since, i³ = -i ]

=−1i×ii

=−ii2=−i−1 = i   [ Since, i² = -1 ]

Q.4: Express the following complex number in x + iy form; 5(6 + 6i) + i(6 + 6i)

Sol:

5(6 + 6i) + i(6 + 6i) = 30 + 30i + 6i + 6i²

= 30 + 36i + 6(-1)    [ Since, i² = -1 ]

= 30 – 6 + 36i

= 24 + 36i

Q.5: Express the following complex number in x + iy form; (2 – 3i) – (-4 + 5i)

Sol:

(2 – 3i) – (-4 + 5i) = 2 – 3i + 4 – 5i

= 6 – 8i

Q.6: Express the following complex number in x + iy form;

(14+i34)−(6+i43)

Sol:

(14+i34)−(6+i43)=14+i34–6–i43

=(14–6)+i(34–43)

=−234+i(−712)

=−234–i712

Q.7: Express the following complex number in x + iy form:

[(13+i73)+(4+i13)]−(−43+i)

Sol:

[(13+i73)+(4+i13)]−(−43+i)

=13+i73+4+i13+43–i

=(13+43+4)+i(73+13–1)

=173+i53

Q.8: Express the following complex number in x + iy form: (1 – i)⁴

Sol:

(1 – i)⁴ = [(1–i)2]2

=[1–2i+i2]2=[1–1–2i]2=(−2i)2=(−2i)×(−2i)=4i2          [ Since i² = -1 ]

= -4

Q.9: Express the following complex number in ‘x + iy’ form; (13+3i)3

Sol:

=(13+3i)3:

=(13)3+(3i)3+3(13)(3i)(13+3i)

=127+27i3+3i(13+3i)

=127+27i3+i+9i2

=127–27i+i–9      [Since, i³ = -i and i² = -1]

=(127–9)+i(−27i+1) = −24227−26i

Q.10: Express the following complex number in ‘x + iy’ form: (−2–i13)3

SoL:

(−2–i13)3=(−1)3(2+i13)3 = −[23+(i3)3+3(2)(i3)(2+i3)] = −[8+i327+2i(2+i3)]

=−[8+i327+4i+2i23)] = −[8–i27+4i–23]      [ Since, i³ = -i and i² = -1 ]

=[223+i10727]=−221–i10727

Q.11: Find the multiplicative inverse of the given complex number, 7 – 6i.

Sol:

Assuming, z = 7 – 6i

Now, z¯¯¯=7+6i

And, |z|² = (7)2 + (-6)2 = 49 + 36 = 85

Therefore, the multiplicative inverse of the given complex number is given by:

z−1=z¯¯¯|z|2=7+6i85=785+685i

Q.12: Find the multiplicative inverse of the given complex number, 7–√+4i

Sol:

Assuming, z = 7–√+4i

Now, z¯¯¯=7–√+4i

And, |z|² = (7–√)2+(4)2=23

Therefore, the multiplicative inverse of the given complex number is given by:

z−1=z¯¯¯|z|2=7√–4i23=7√23–423i

Q.13: Find the multiplicative inverse of the given complex number, – i

Sol:

Assuming, z = – i

Now, z¯¯¯=i

And, |z|² = (-i)² = 1

Therefore, the multiplicative inverse of the given complex number is given by:

z−1=z¯¯¯|z|2=i1 = i

Q.14: Express the following complex number in ‘x + iy’ form: (3+i5√)(3–i5√)(3√+i2√)–(3√–i2√)

Sol:

(3+i5√)(3–i5√)(3√+i2√)–(3√–i2√)=(3)2–(i5√)23√+i2√–3√+i2√

Now, by using (a + b) (a – b) = (a – b)²

=9–5i222√i=9–5×(−1)22√i=9+522√i×ii=14i22√i2

=14i−22√=−7i2√×2√2√=−72√i2

Exercise: 5.2

Q.1: Find out the modulus and argument of the given complex number:

z=−1−i3–√

Sol:

Assuming rcosΘ=−1andrsinΘ=−3–√

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(−1)2+(−3–√)2⇒r2(cos2Θ+sin2Θ)=1+3⇒r2(1)=4⇒r=4–√=2

Therefore, Modulus = 2   [ Since, r > 0 ]

Now, 2cosΘ=−1and2sinΘ=3–√⇒cosΘ=−12andsinΘ=3√2

Since, the values of both cosΘandsinΘare negative and they both are negative in 3^rd quadrant

Therefore, Argument = −(π–π3)=−2π3

Q.2: Find out the modulus and argument of the given complex number

z=−3–√+i

Sol:

Assuming rcosΘ=−3–√andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(−3–√)2+(1)2⇒r2(cos2Θ+sin2Θ)=3+1⇒r2(1)=4⇒r=4–√=2

Therefore, Modulus = 2              [ Since, r > 0 ]

Now, 2cosΘ=−3–√and2sinΘ=1⇒cosΘ=−3√2andsinΘ=12

Here, Θ lies in the 2^nd quadrant.

Therefore, Argument = (π–π6)=5π6

Q.3: Convert the following complex number in polar form: 1 – i

Sol:

Assuming rcosΘ=1andrsinΘ=−1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1)2+(−1)2⇒r2(cos2Θ+sin2Θ)=1+1⇒r2(1)=2⇒r=2–√

Therefore, Modulus = 2–√         [ Since, r > 0 ]

Now, 2–√cosΘ=1and2–√sinΘ=−1

⇒cosΘ=12√andsinΘ=−12√

Here, Θ lies in 4^th quadrant.

Therefore, Θ = (−π4)

Therefore, the required polar form is:

1–i=rcosΘ+irsinΘ=2–√cos(−π4)+i2–√sin(−π4)

Therefore, 1−i=2−−√[cos(−π4)+isin(−π4)]

Q.4: Convert the following complex number in polar form;   -1 + i

Sol:

Assuming rcosΘ=−1andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(−1)2+(1)2⇒r2(cos2Θ+sin2Θ)=1+1⇒r2(1)=2⇒r=2–√

Therefore, Modulus =  2–√               [ Since, r > 0 ]

Now,2–√cosΘ=−1and2–√sinΘ=1

⇒cosΘ=−12√andsinΘ=12√

Here, Θ lies in 2^nd quadrant.

Therefore, Θ = (π−π4)=3π4

So, the required polar form is:

−1+i=rcosΘ+irsinΘ=2–√cos(3π4)+i2–√sin(3π4)

Therefore, -1 + i = 2−−√[cos(3π4)+isin(3π4)]

Q.5: Convert the following complex number in polar form;  -3

Sol:

Assuming rcosΘ=−3andrsinΘ=0

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(−3)2+(0)2⇒r2(cos2Θ+sin2Θ)=9⇒r2(1)=9⇒r=9–√=3

Therefore, Modulus = 3    [ Since, r > 0]

Now, 3cosΘ=−3and3sinΘ=0⇒cosΘ=−1andsinΘ=0

Therefore, Θ=π

So, the required polar form is:

−3=rcosΘ+irsinΘ=3cos(π)+i.3sin(π)

Therefore, −3=3[cos(π)+isin(π)]

Q.6: Convert the following complex number in polar form;   -1 – i

Sol:

Assuming rcosΘ=−1andrsinΘ=−1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(−1)2+(−1)2⇒r2(cos2Θ+sin2Θ)=1+1⇒r2(1)=2⇒r=2–√

Therefore, Modulus = 2–√           [ Since, r > 0 ]

Now, 2–√cosΘ=−1and2–√sinΘ=−1

⇒cosΘ=−12√andsinΘ=−12√

Q.7: Convert the following complex number in polar form: z=3–√+i

Sol:

z=3–√+i

Assuming rcosΘ=3–√andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(3–√)2+(1)2⇒r2(cos2Θ+sin2Θ)=3+1⇒r2(1)=4⇒r=4–√=2

Therefore, Modulus = 2     [Since, r > 0]

Now,

2cosΘ=3–√and2sinΘ=1⇒cosΘ=3√2andsinΘ=12

Here, Θ lies in 1^st quadrant.

Therefore, Θ = (π6)

So, the required polar form is:

3–√+i=rcosΘ+irsinΘ=2cos(π6)+i2sin(π6)

Therefore, 3−−√+i=2[cos(π6)+isin(π6)]

Q.8: Convert the following complex number in polar form: i

Sol:

Assuming rcosΘ=0andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(0)2+(1)2⇒r2(cos2Θ+sin2Θ)=1⇒r2(1)=1⇒r=1–√=1

Therefore, Modulus = 1     [ Since, as r > 0)]

Now, cosΘ=0and3sinΘ=1

Therefore, Θ=π2

So, the required polar form is:

i=cos(π2)+isin(π2)

Miscellaneous Exercise

Q-1: Evaluate the following:

[i18+(1i)25]3

Sol:

[i18+(1i)25]3  =  [i4×4+2+1i4×6+1]3  =  [(i4)4⋅i2+1(i6)4⋅i]3  =  [i2+1i]3   [Since, i⁴ = 1]

=[−1+1i×ii]3=[−1+ii2]3=[−1–i]3

=(−1)3[1+i]3=−[1+i3+3i(i+1)]

=−[1–i3+3i+3i2]

= – [1 – i + 3i – 3] = – [-2 + 2i]

= 2 – 2i

Q-2: For the two given complex number z₁ and z₂ below prove that

Re(z₁z₂) = Re z₁Re z₂ – Im z₁ Im z₂

Sol:

Assuming z₁ = a₁ + ib₁ and z₂ = a₂ + ib₂

Therefore, z₁z₂ = (a₁ + ib₁) ( a₂ + ib₂)

= a₁(a₂ + ib₂) + ib₂(a₂ + ib₂)

= a₁a₂ + ia₁b₂ + ib₁a₂ + i²b₁b₂

= a₁a₂ + ia₁b₂ + ib₁a₂ – b₁b₂

= (a₁a₂ – b₁b₂) + i(a₁b₂ + b₁a₂)

Re(z₁z₂) = a₁a₂ – b₁b₂

Re(z₁z₂) = Rez₁ Rez₂  – Imz₁ Imz₂

Hence, Proved

Q-3: Express (11–4i–21+i)(3–4i5+i) in the standard form i.e. a + ib.

Sol:

[11–4i–21+i][3–4i5+i]=[(1+i)−2(1–4i)(1−4i)(1+i)][3–4i5+i] =[1+i–2+8i1+i–4i−4i2][3–4i5+i]=[−1+9i5–3i][3–4i5+i] =[−3+4i+27i–36i225+5i–15i–3i2]=33+31i28–10i=33+31i2(14–5i) =33+31i2(14–5i)×14+5i14+5i

Now, on multiplying and dividing by (14 + 5i) we will get:

=462+165i+434i+155i22[(14)2–(5i)2]=307+599i2(196–25i2) =307+599i2(221)=307+599i442 =307442+i599442

Q-4: If a−ib=x−iyu−iv−−−−−√ then prove that (a2+b2)2=x2+y2u2+v2

Sol:

a–ib=x−iyu−iv−−−−−√ =x–iyu–iv×x+iyu+iv−−−−−−−−−√

Now, on multiplying and dividing by (u + iv) we will get:

=(xu+yv)+i(xv−yu)u2+v2−−−−−−−−−−−−−−−√

Therefore, (a–ib)2=(xu+yv)+i(xv–yu)u2+v2

⇒a2–2iab–b2=(xu+yv)+i(xv−yu)u2+v2

On comparing both sides, we will get:

a2–b2=xu+yvu2+v2and−2ab=xv−yuu2+v2……………..(a)

(a² + b²)² = (a² – b²)² + 4a²b²

=(xu+yvu2+v2)2+(xv−uu2+v2)2

Now, by using equation (a) we will get:

=x2u2+y2v2+2xyuv+x2v2+y2u2−2xyuv(u2+v2)2 =x2u2+y2v2+x2v2+y2u2(u2+v2)2 =x2(u2+v2)+y2(u2+v2)2(u2+v2)2 =(x2+y2)(u2+v2)(u2+v2)2=x2+y2u2+v2

Hence, proved

Q-5) Convert the following complex number in polar form:

1: 1+7i(2–i)2

2:1+3i1–2i

Sol:

1: Here, z = 1+7i(2–i)2

1+7i(2–i)2=1+7i4+i2–4i=1+7i4–1–4i

=1+7i3–4i×3+4i3+4i=3+4i+21i+28i232+42

=3+4i+21i–2832+42=−25+25i25 = -i + 1

Assuming rcosΘ=−1andrsinΘ=1

On squaring on both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(−1)2+(1)2⇒r2(cos2Θ+sin2Θ)=1+1⇒r2(1)=2⇒r=2–√

Therefore, Modulus = 2–√        [ Since, r > 0]

Now, 2–√cosΘ=−1and2–√sinΘ=1⇒cosΘ=−12√andsinΘ=12√

Here, Θ lies in 2^nd quadrant.

Therefore, Θ = (π−π4)=3π4

Therefore, the required polar form is:

−1+i=rcosΘ+irsinΘ=2–√cos(3π4)+i2–√sin(3π4)

=2–√[cos(3π4)+isin(3π4)]

2: Here, z = 1+3i1−2i

1+3i1−2i=1+3i1−2i×1+2i1+2i

=1+2i+3i–61+4=−5+5i5 = -i + 1

Assuming rcosΘ=−1andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(−1)2+(1)2⇒r2(cos2Θ+sin2Θ)=1+1⇒r2(1)=2⇒r=2–√

Therefore, Modulus = 2–√        [ Since, r > 0]

Now,2–√cosΘ=−1and2–√sinΘ=1

⇒cosΘ=−12√andsinΘ=12√

Here, Θ lies in 2^nd quadrant.

Now, Θ = (π−π4)=3π4

Therefore, the required polar form is:

−1+i=rcosΘ+irsinΘ=2–√cos(3π4)+i2–√sin(3π4)

=2–√[cos(3π4)+isin(3π4)]

Q-6: Solve the equation given below:

3y² – 4y + 203 = 0

Sol:

Here, 3y² – 4y + 203 = 0

The given equation can be also be written as,

9y² – 12y + 20 = 0

On comparing the given equation with ax² + bx + c = 0, we will get:

a = 9, b = -12 and c = 20

Now, the discriminant is:

D=b2–4ac=(−12)2–4×9×20=144–720=−576

Now, the solution is:

y=−b±D√2a =−(12)±−576√2(9) =12±576i2√18

=12±24i18=2±4i3

=23±43i

Therefore, y = =23±43i

Q-7: Solve the equation given below:

y² – 2y + 32 = 0

Sol:

Here, y² – 2y + 32= 0

The given equation can be also be written as:

2y² – 4y + 3 = 0

On comparing the given equation with ax² + bx + c = 0, we will get:

a = 2, b = -4 and c = 3

Now, the discriminant is:

D=b2–4ac=(−4)2–4×(2)×(3)=16–24=−8

Now, the solution is:

y=−b±D√2a =−(−4)±−8√2(2) =4±8i2√4

=4±22√i4=2±2√i2

=1±2√2i

Therefore, y =1±2√2i

Q-8: Solve the equation given below:

27y² – 10y + 1 = 0

Sol:

Here, 27y² – 10y + 1 = 0

On comparing the given equation with ax² + bx + c = 0, we will get:

a = 27, b = -10 and c = 1

Now, the discriminant is:

D=b2–4ac=(−10)2–4×(27)×(1)=100–108=−8

Now, the solution is:

y=−b±D√2a =−(−10)±−8√2(27) =10±8i2√54

=10±22√i54=5±2√i27

=57±2√27i

Therefore, y =57±2√27i

Q-9: Solve the equation given below:

21y² – 28y + 10 = 0

Sol:

Here, 21y² – 28y + 10 = 0

On comparing the given equation with ax² + bx + c = 0, we will get:

a = 21, b = -28 and c = 10

Now, the discriminant is:

D=b2–4ac=(−28)2–4×(21)×(10)=784–840=−56

Now, the solution is:

Since, y=−b±D√2a

=−(−28)±−56√2(21) =28±56i2√42

=28±214√i42=14±14√i21=32±14√21i

Therefore, y =32±14√21i

Q-10:  z₁ = 2 – i, z₂ = 1 + i, find ∣∣z1+z2+1z1–z2+i∣∣

Sol:

Here, z₁ = 2 – i, z₂ = 1 + i

∣∣z1+z2+1z1–z2+i∣∣=∣∣(2–i)+(1+i)+1(2–i)–(1+i)+1∣∣

=∣∣42–2i∣∣=∣∣21–i∣∣=∣∣21–i×1+i1+i∣∣=∣∣2(1+i)1–i2∣∣

=∣∣2(1+i)1+1∣∣=∣∣2(1+i)2∣∣=|1+i|=12+12−−−−−−√=2–√

Q-11:  If x + iy = (a+i)22a2+1 then prove that

x² + y² =  (a2+1)2(2a+1)2

Sol:

Here, x + iy = (a+i)22a2+1

=a2+i2+2ai2a2+1=a2–1+2ai2a2+1=a2–12a2+1+i2a2a2+1

On comparing both sides, we will get:

x=a2–12a2+1andy=2a2a2+1

Therefore, x2+y2=(a2–12a2+1)2+(2a2a2+1)2

=a4+1–2a2+4a2(2a+1)2=a4+1+2a2(2a2+1)2 =(a2+1)2(2a2+1)2

Hence, proved

Q-12: If z₁ = 2 – i and z₂ = -2 + i is given find out,

1: Im(1z1z1¯¯¯¯¯)   

2:Re(z1z2z1)

Sol:

Here, z₁ = 2 – i and z₂ = -2 + i

1. Im(1z1z1¯¯¯¯¯)

=1(2+i)(−2+i)=122–i2=14+1=15

On comparing both sides, we will get:

Im(1z1z1¯¯¯¯¯)=0

2. Re(z1z2z1)

z₁z₂ = (2 – i)(- 2 + i) = -4 + 2i + 2i – i² = -4 + 4i – (-1) = -3 + 4i

z1¯¯¯¯¯=2+i z1z2z1¯¯¯¯¯=−3+4i2+i

Now multiplying and dividing it by (2 – i), we will get:

=−3+4i2+i×2–i2–i

=−6+3i+8i–4i222–i2

=−6+11i–4(−1)4+1

=2+11i5=−25+115i

On comparing both the sides, we will get:

Re(z1z2z1¯¯¯¯¯)=−25

Q-13: Find out the modulus and argument of the given complex number

1+2i1−3i

Sol:

Here, z = 1+2i1–3i

1+2i1−3i×1+3i1+3i=1+3i+2i+6i212−9i2

=1+3i+2i+6i212+9=−5+5i10

=−510+510i=−12+12i

Assuming rcosΘ=−1/2andrsinΘ=1/2

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(−1/2)2+(1/2)2⇒r2(cos2Θ+sin2Θ)=1/4+1/4⇒r2(1)=1/2⇒r=12√

Therefore, Modulus = 12√          [ Since, r > 0]

Now, 12√cosΘ=−12and12√sinΘ=12

⇒cosΘ=−12√andsinΘ=12√

Here, Θ lies in 2^nd  quadrant.

Therefore, Θ=(π−π4)=3π4

Q-14: Find the real number a and b if (3 + 5i) (a – ib) is the conjugate if -6 – 24i.

Sol:

Assuming z = (3 + 5i) (a – ib)

z = 3a + 5ai – 3bi – 5bi² = 3a + 5ai – 3bi + 5b = (3a + 5b) + i(5a – 3b)

Therefore, z¯¯¯=(3a+5b)–i(5a–3b)

Here, given that z¯¯¯=−6–24i

Therefore, (3a + 5b) – i(5a – 3b) = -6 – 24i

On comparing both sides, we will get:

3a + 5b = -6 . . . . . . . . . . . . (a)

5a – 3b = 24 . . . . . . . . . . . . (b)

On solving these two equations, we will get:

9a + 15b = -18

25a – 15b = 120

34a = 102

Therefore, a = 3

Now, 3(3) + 5b = -6       [From equation (a)]

5b = -15

Therefore, b = -3

Hence, a = 3 and b = -3

Q-15: Find the modulus of 1+i1–i−1−i1+i

Sol:

1+i1−i–1−i1+i=(1+i)2–(1–i)2(1−i)(1+i)

=1+i2+2i−1−i2+2i1+1=4i2=2i

⇒   ∣∣1+i1−i−1−i1+i∣∣=|2i|=22−−√=2

Q-16: If (a + ib)³ = u + iv, then prove that:

ua+vb=4(a2–b2)

Sol:

Here, (a + ib)³ = u + iv

a³ + (ib)³ + 3a(ib)(a + ib) = u + iv

a³ + i³b³ + 3a²bi + 3ab²i² = u + iv

a³ – ib³ + 3a²bi – 3ab² = u + iv

(a³ – 3ab²) + i(3a²b – b³) = u + iv

On comparing both the sides, we will get:

u = a³ – 3ab²     and      v = 3a²b – b³

Hence, Proved

Q-17: If A and B are two different complex numbers with |B| = 1, then find ∣∣B–A1–A¯¯¯¯B∣∣

Sol:

Assuming, A = x + iy and B = u + iv

Here, |B| = 1,

Therefore, u2+v2−−−−−−√=1

⇒ u² + v² = 1 . . . . . . . . . . . . (a)

∣∣B–A1–A¯¯¯¯B∣∣=∣∣(x+iy)−(u+iv)1–(u–iv)(x+iy)∣∣

=∣∣(x−u)+i(y−v)1−(xu+uiy+ivx+yv)∣∣

=∣∣(x–u)+i(y−v)(1−xu–yv)+i(xv−yu)∣∣=|(x−v)+i(y−v)||(1−xu−yv)+i(xv–yu)|

Since, ∣∣z1z2∣∣=|z1||z2|

Therefore, =(x−u)2+(y–v)2√(1−xu−yv)+i(xv−uy)√

=x2+u2−2ux+y2+v2−2vy√1+u2x2+v2y2−2ux+2uvxy–2vy+v2x2+u2y2−2uvxy√

=(x2+y2)+u2+v2−2ux–2vy√1+u2(x2+y2)+v2(x2+y2)−2xu−2yv√

=1+u2+v2–2xu–2yv√1+u2+v2−2xu−2yv√=1                 [ From equation (a) ]

Therefore, ∣∣B−A1−A¯¯¯¯B∣∣=1

Q-18: Find the number of non-zero solutions of the equation |1 – i|^y = 2^y

Sol:

|1 – i|^y = 2^y

⇒(12+(−1)2−−−−−−−−−√)y=2y

⇒   (2–√)y=2y   ⇒2y2=y

⇒   y=2y⇒2y−y=0⇒y=0

Therefore, there is only one integral solution 0.

Q.19: If (s + it) (u + iv) (w + ix) (y + iz) = X + iY, then prove that:

(s² + t²) (u² + v²) (w² + x²) (y² + z²) = X² + Y².

Sol:

Here, (s + it) (u + iv) (w + ix) (y + iz) = X + iY

Therefore, |(s + it)(u + iv)(w + ix)(y + iz)| = |X + iY|

|(s + it)| × |(u + iv)| × |(w + ix)| × |(y + iz)| = |X + iY|  [because |z₁z₂| = |z₁||z₂|]

⇒s2+t2−−−−−−√×u2+v2−−−−−−√×w2+x2−−−−−−√×y2+z2−−−−−−√=X2+Y2−−−−−−−√

Now, on squaring both the sides, we will get:

(s² + t²) (u² + v²) (w² + x²) (y² + z²) = X² + Y²

Hence, Proved

Q.20: Find the least positive value of x for the following:

(1+i1–i)x=1

Sol:

(1+i1−i)x=1 ⇒(1+i1−i×1+i1+i)x=1 ⇒((1+i)21+1)x=1

⇒   (1+i2+2i2)x=1

⇒(2i2)x=1⇒ix=1

Hence, x = 4m, where m belongs to Z.

Thus, the least positive integer = 1

Therefore, the least positive integral for given problem = x = 4m = 4 × 1 = 4