NCERT Solution for Class 11 Science Mathematics 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 3rd 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 2nd 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 4th 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 2nd 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 1st 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 2nd 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 2nd 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 2nd  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 = 2y



Sol:

|1 – i|y = 2y

⇒(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