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Complex Numbers and Quadratic Equations

Jharkhand Board · Class 11 · Mathematics

NCERT Solutions for Complex Numbers and Quadratic Equations — Jharkhand Board Class 11 Mathematics.

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28 Questions Solved · 2 Sections

Exercise 4.1

1Express (5i)(−35i)(5i)\left(-\dfrac{3}{5}i\right) in the form a+iba + ib.Show solution
Given: (5i)(−35i)(5i)\left(-\dfrac{3}{5}i\right)

Working:
(5i)(−35i)=5×(−35)×i×i=−3×i2
(5i)\left(-\frac{3}{5}i\right) = 5 \times \left(-\frac{3}{5}\right) \times i \times i = -3 \times i^2

Since i2=−1i^2 = -1:
=−3×(−1)=3
= -3 \times (-1) = 3


Answer: 3+0i3 + 0i, i.e., a=3, b=0a = 3,\ b = 0.
2Express i9+i19i^9 + i^{19} in the form a+iba + ib.Show solution
Given: i9+i19i^9 + i^{19}

Concept: For any integer nn, the powers of ii cycle with period 4: i1=i, i2=−1, i3=−i, i4=1i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1.

Working:
i9=i4×2+1=(i4)2⋅i=1⋅i=i
i^9 = i^{4\times2+1} = (i^4)^2 \cdot i = 1 \cdot i = i

i19=i4×4+3=(i4)4⋅i3=1⋅(−i)=−i
i^{19} = i^{4\times4+3} = (i^4)^4 \cdot i^3 = 1 \cdot (-i) = -i

i9+i19=i+(−i)=0
i^9 + i^{19} = i + (-i) = 0


Answer: 0+0i0 + 0i, i.e., a=0, b=0a = 0,\ b = 0.
3Express i−39i^{-39} in the form a+iba + ib.Show solution
Given: i−39i^{-39}

Working:
i−39=1i39
i^{-39} = \frac{1}{i^{39}}

Now, 39=4×9+339 = 4 \times 9 + 3, so i39=i3=−ii^{39} = i^3 = -i.
i−39=1−i=1−i×ii=i−i2=i−(−1)=i1=i
i^{-39} = \frac{1}{-i} = \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i


Answer: 0+i0 + i, i.e., a=0, b=1a = 0,\ b = 1.
4Express 3(7+i7)+i(7+i7)3(7 + i7) + i(7 + i7) in the form a+iba + ib.Show solution
Given: 3(7+i7)+i(7+i7)3(7 + i7) + i(7 + i7)

Working:
=21+21i+7i+7i2
= 21 + 21i + 7i + 7i^2

=21+28i+7(−1)
= 21 + 28i + 7(-1)

=21−7+28i=14+28i
= 21 - 7 + 28i = 14 + 28i


Answer: 14+28i14 + 28i, i.e., a=14, b=28a = 14,\ b = 28.
5Express (1−i)−(−1+i6)(1 - i) - (-1 + i6) in the form a+iba + ib.Show solution
Given: (1−i)−(−1+6i)(1 - i) - (-1 + 6i)

Working:
=1−i+1−6i=2−7i
= 1 - i + 1 - 6i = 2 - 7i


Answer: 2−7i2 - 7i, i.e., a=2, b=−7a = 2,\ b = -7.
6Express (15+i25)−(4+i52)\left(\dfrac{1}{5} + i\dfrac{2}{5}\right) - \left(4 + i\dfrac{5}{2}\right) in the form a+iba + ib.Show solution
Given: (15+25i)−(4+52i)\left(\dfrac{1}{5} + \dfrac{2}{5}i\right) - \left(4 + \dfrac{5}{2}i\right)

Working:
=15−4+(25−52)i
= \frac{1}{5} - 4 + \left(\frac{2}{5} - \frac{5}{2}\right)i

=1−205+(4−2510)i
= \frac{1 - 20}{5} + \left(\frac{4 - 25}{10}\right)i

=−195−2110i
= -\frac{19}{5} - \frac{21}{10}i


Answer: −195−2110i-\dfrac{19}{5} - \dfrac{21}{10}i, i.e., a=−195, b=−2110a = -\dfrac{19}{5},\ b = -\dfrac{21}{10}.
7Express [(13+i73)+(4+i13)]−(−43+i)\left[\left(\dfrac{1}{3} + i\dfrac{7}{3}\right) + \left(4 + i\dfrac{1}{3}\right)\right] - \left(-\dfrac{4}{3} + i\right) in the form a+iba + ib.Show solution
Given: [(13+73i)+(4+13i)]−(−43+i)\left[\left(\dfrac{1}{3} + \dfrac{7}{3}i\right) + \left(4 + \dfrac{1}{3}i\right)\right] - \left(-\dfrac{4}{3} + i\right)

Step 1: Add the first two complex numbers:
13+4+(73+13)i=1+123+83i=133+83i
\frac{1}{3} + 4 + \left(\frac{7}{3} + \frac{1}{3}\right)i = \frac{1+12}{3} + \frac{8}{3}i = \frac{13}{3} + \frac{8}{3}i


Step 2: Subtract the third:
133−(−43)+(83−1)i=13+43+8−33i=173+53i
\frac{13}{3} - \left(-\frac{4}{3}\right) + \left(\frac{8}{3} - 1\right)i = \frac{13+4}{3} + \frac{8-3}{3}i = \frac{17}{3} + \frac{5}{3}i


Answer: 173+53i\dfrac{17}{3} + \dfrac{5}{3}i, i.e., a=173, b=53a = \dfrac{17}{3},\ b = \dfrac{5}{3}.
8Express (1−i)4(1 - i)^4 in the form a+iba + ib.Show solution
Given: (1−i)4(1-i)^4

Working:
(1−i)2=1−2i+i2=1−2i−1=−2i
(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i

(1−i)4=[(1−i)2]2=(−2i)2=4i2=4(−1)=−4
(1-i)^4 = \left[(1-i)^2\right]^2 = (-2i)^2 = 4i^2 = 4(-1) = -4


Answer: −4+0i-4 + 0i, i.e., a=−4, b=0a = -4,\ b = 0.
9Express (13+3i)3\left(\dfrac{1}{3} + 3i\right)^3 in the form a+iba + ib.Show solution
Given: (13+3i)3\left(\dfrac{1}{3} + 3i\right)^3

Concept: Use the identity (a+b)3=a3+3a2b+3ab2+b3(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 with a=13a = \dfrac{1}{3} and b=3ib = 3i.

Working:
=(13)3+3(13)2(3i)+3(13)(3i)2+(3i)3
= \left(\frac{1}{3}\right)^3 + 3\left(\frac{1}{3}\right)^2(3i) + 3\left(\frac{1}{3}\right)(3i)^2 + (3i)^3

=127+3⋅19⋅3i+3⋅13⋅9i2+27i3
= \frac{1}{27} + 3 \cdot \frac{1}{9} \cdot 3i + 3 \cdot \frac{1}{3} \cdot 9i^2 + 27i^3

=127+i+9(−1)+27(−i)
= \frac{1}{27} + i + 9(-1) + 27(-i)

=127−9+(1−27)i
= \frac{1}{27} - 9 + (1 - 27)i

=1−24327−26i=−24227−26i
= \frac{1 - 243}{27} - 26i = -\frac{242}{27} - 26i


Answer: −24227−26i-\dfrac{242}{27} - 26i, i.e., a=−24227, b=−26a = -\dfrac{242}{27},\ b = -26.
10Express (−2−13i)3\left(-2 - \dfrac{1}{3}i\right)^3 in the form a+iba + ib.Show solution
Given: (−2−13i)3\left(-2 - \dfrac{1}{3}i\right)^3

Concept: Use (a+b)3=a3+3a2b+3ab2+b3(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 with a=−2a = -2 and b=−13ib = -\dfrac{1}{3}i.

Working:
=(−2)3+3(−2)2(−i3)+3(−2)(−i3)2+(−i3)3
= (-2)^3 + 3(-2)^2\left(-\frac{i}{3}\right) + 3(-2)\left(-\frac{i}{3}\right)^2 + \left(-\frac{i}{3}\right)^3

=−8+3(4)(−i3)+(−6)(i29)+(−i327)
= -8 + 3(4)\left(-\frac{i}{3}\right) + (-6)\left(\frac{i^2}{9}\right) + \left(-\frac{i^3}{27}\right)

=−8−4i+(−6)(−19)+(−−i27)
= -8 - 4i + (-6)\left(-\frac{1}{9}\right) + \left(-\frac{-i}{27}\right)

=−8−4i+23+i27
= -8 - 4i + \frac{2}{3} + \frac{i}{27}

=(−8+23)+(−4+127)i
= \left(-8 + \frac{2}{3}\right) + \left(-4 + \frac{1}{27}\right)i

=−24+23+−108+127i
= \frac{-24+2}{3} + \frac{-108+1}{27}i

=−223−10727i
= -\frac{22}{3} - \frac{107}{27}i


Answer: −223−10727i-\dfrac{22}{3} - \dfrac{107}{27}i, i.e., a=−223, b=−10727a = -\dfrac{22}{3},\ b = -\dfrac{107}{27}.
11Find the multiplicative inverse of 4−3i4 - 3i.Show solution
Given: z=4−3iz = 4 - 3i

Formula: z−1=zˉ∣z∣2z^{-1} = \dfrac{\bar{z}}{|z|^2}

Working:
zˉ=4+3i,∣z∣2=42+(−3)2=16+9=25
\bar{z} = 4 + 3i, \quad |z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25

z−1=4+3i25=425+325i
z^{-1} = \frac{4 + 3i}{25} = \frac{4}{25} + \frac{3}{25}i


Answer: The multiplicative inverse of 4−3i4-3i is 425+325i\dfrac{4}{25} + \dfrac{3}{25}i.
12Find the multiplicative inverse of 5+3i\sqrt{5} + 3i.Show solution
Given: z=5+3iz = \sqrt{5} + 3i

Formula: z−1=zˉ∣z∣2z^{-1} = \dfrac{\bar{z}}{|z|^2}

Working:
zˉ=5−3i,∣z∣2=(5)2+32=5+9=14
\bar{z} = \sqrt{5} - 3i, \quad |z|^2 = (\sqrt{5})^2 + 3^2 = 5 + 9 = 14

z−1=5−3i14=514−314i
z^{-1} = \frac{\sqrt{5} - 3i}{14} = \frac{\sqrt{5}}{14} - \frac{3}{14}i


Answer: The multiplicative inverse of 5+3i\sqrt{5}+3i is 514−314i\dfrac{\sqrt{5}}{14} - \dfrac{3}{14}i.
13Find the multiplicative inverse of −i-i.Show solution
Given: z=−i=0−iz = -i = 0 - i

Formula: z−1=zˉ∣z∣2z^{-1} = \dfrac{\bar{z}}{|z|^2}

Working:
zˉ=0+i=i,∣z∣2=02+(−1)2=1
\bar{z} = 0 + i = i, \quad |z|^2 = 0^2 + (-1)^2 = 1

z−1=i1=i
z^{-1} = \frac{i}{1} = i


Verification: (−i)(i)=−i2=−(−1)=1(-i)(i) = -i^2 = -(-1) = 1 ✓

Answer: The multiplicative inverse of −i-i is ii (i.e., 0+1⋅i0 + 1\cdot i).
14Express (3+i5)(3−i5)(3+2 i)−(3−i2)\dfrac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + \sqrt{2}\,i) - (\sqrt{3} - i\sqrt{2})} in the form a+iba + ib.Show solution
Given: (3+i5)(3−i5)(3+2 i)−(3−i2)\dfrac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + \sqrt{2}\,i) - (\sqrt{3} - i\sqrt{2})}

Step 1: Simplify the numerator using (a+ib)(a−ib)=a2+b2(a+ib)(a-ib) = a^2+b^2:
(3+i5)(3−i5)=32+(5)2=9+5=14
(3 + i\sqrt{5})(3 - i\sqrt{5}) = 3^2 + (\sqrt{5})^2 = 9 + 5 = 14


Step 2: Simplify the denominator:
(3+2 i)−(3−2 i)=3+2 i−3+2 i=22 i
(\sqrt{3} + \sqrt{2}\,i) - (\sqrt{3} - \sqrt{2}\,i) = \sqrt{3} + \sqrt{2}\,i - \sqrt{3} + \sqrt{2}\,i = 2\sqrt{2}\,i


Step 3: Divide:
1422 i=72 i=72 i×−i−i=−7i2 (−i2)=−7i2=0−72i
\frac{14}{2\sqrt{2}\,i} = \frac{7}{\sqrt{2}\,i} = \frac{7}{\sqrt{2}\,i} \times \frac{-i}{-i} = \frac{-7i}{\sqrt{2}\,(-i^2)} = \frac{-7i}{\sqrt{2}} = 0 - \frac{7}{\sqrt{2}}i

Rationalising: 72=722\dfrac{7}{\sqrt{2}} = \dfrac{7\sqrt{2}}{2}

=0−722i
= 0 - \frac{7\sqrt{2}}{2}i


Answer: 0−722i0 - \dfrac{7\sqrt{2}}{2}i, i.e., a=0, b=−722a = 0,\ b = -\dfrac{7\sqrt{2}}{2}.

Miscellaneous Exercise on Chapter 4

1Evaluate: [i18+(1i)25]3\left[i^{18} + \left(\dfrac{1}{i}\right)^{25}\right]^3.Show solution
Given: [i18+(1i)25]3\left[i^{18} + \left(\dfrac{1}{i}\right)^{25}\right]^3

Step 1: Simplify i18i^{18}:
i18=i4×4+2=(i4)4⋅i2=1⋅(−1)=−1
i^{18} = i^{4\times4+2} = (i^4)^4 \cdot i^2 = 1 \cdot (-1) = -1


Step 2: Simplify (1i)25=i−25\left(\dfrac{1}{i}\right)^{25} = i^{-25}:
i−25=1i25,i25=i4×6+1=i
i^{-25} = \frac{1}{i^{25}}, \quad i^{25} = i^{4\times6+1} = i

i−25=1i=−i−i2=−i1⋅ii⇒1i×ii=ii2=i−1=−i
i^{-25} = \frac{1}{i} = \frac{-i}{-i^2} = \frac{-i}{1} \cdot \frac{i}{i}\Rightarrow \frac{1}{i}\times\frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i


Step 3: Add and cube:
i18+i−25=−1+(−i)=−1−i
i^{18} + i^{-25} = -1 + (-i) = -1 - i

(−1−i)3=−(1+i)3
(-1-i)^3 = -(1+i)^3

(1+i)2=1+2i+i2=1+2i−1=2i
(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i

(1+i)3=(1+i)(2i)=2i+2i2=2i−2=−2+2i
(1+i)^3 = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i

−(1+i)3=−(−2+2i)=2−2i
-(1+i)^3 = -(-2+2i) = 2 - 2i


Answer: 2−2i2 - 2i.
2For any two complex numbers z1z_1 and z2z_2, prove that Re⁡(z1z2)=Re⁡z1Re⁡z2−Im⁡z1Im⁡z2\operatorname{Re}(z_1 z_2) = \operatorname{Re}z_1\operatorname{Re}z_2 - \operatorname{Im}z_1\operatorname{Im}z_2.Show solution
Given: z1z_1 and z2z_2 are any two complex numbers.

Let z1=a+ibz_1 = a + ib and z2=c+idz_2 = c + id, where a,b,c,d∈Ra,b,c,d \in \mathbb{R}.

So Re⁡z1=a, Im⁡z1=b, Re⁡z2=c, Im⁡z2=d\operatorname{Re}z_1 = a,\ \operatorname{Im}z_1 = b,\ \operatorname{Re}z_2 = c,\ \operatorname{Im}z_2 = d.

Compute z1z2z_1 z_2:
z1z2=(a+ib)(c+id)=ac+iad+ibc+i2bd=(ac−bd)+i(ad+bc)
z_1 z_2 = (a+ib)(c+id) = ac + iad + ibc + i^2 bd = (ac - bd) + i(ad + bc)


Therefore:
Re⁡(z1z2)=ac−bd=Re⁡z1⋅Re⁡z2−Im⁡z1⋅Im⁡z2
\operatorname{Re}(z_1 z_2) = ac - bd = \operatorname{Re}z_1 \cdot \operatorname{Re}z_2 - \operatorname{Im}z_1 \cdot \operatorname{Im}z_2


Hence proved.
3Reduce (11−4i−21+i)(3−4i5+i)\left(\dfrac{1}{1-4i} - \dfrac{2}{1+i}\right)\left(\dfrac{3-4i}{5+i}\right) to the standard form.Show solution
Step 1: Simplify 11−4i\dfrac{1}{1-4i}:
11−4i=1+4i(1−4i)(1+4i)=1+4i1+16=1+4i17
\frac{1}{1-4i} = \frac{1+4i}{(1-4i)(1+4i)} = \frac{1+4i}{1+16} = \frac{1+4i}{17}


Step 2: Simplify 21+i\dfrac{2}{1+i}:
21+i=2(1−i)(1+i)(1−i)=2(1−i)2=1−i
\frac{2}{1+i} = \frac{2(1-i)}{(1+i)(1-i)} = \frac{2(1-i)}{2} = 1-i


Step 3: Subtract:
1+4i17−(1−i)=1+4i−17(1−i)17=1+4i−17+17i17=−16+21i17
\frac{1+4i}{17} - (1-i) = \frac{1+4i - 17(1-i)}{17} = \frac{1+4i-17+17i}{17} = \frac{-16+21i}{17}


Step 4: Simplify 3−4i5+i\dfrac{3-4i}{5+i}:
3−4i5+i=(3−4i)(5−i)(5+i)(5−i)=15−3i−20i+4i225+1=15−23i−426=11−23i26
\frac{3-4i}{5+i} = \frac{(3-4i)(5-i)}{(5+i)(5-i)} = \frac{15-3i-20i+4i^2}{25+1} = \frac{15-23i-4}{26} = \frac{11-23i}{26}


Step 5: Multiply:
−16+21i17×11−23i26=(−16+21i)(11−23i)442
\frac{-16+21i}{17} \times \frac{11-23i}{26} = \frac{(-16+21i)(11-23i)}{442}

(−16+21i)(11−23i)=−176+368i+231i−483i2=−176+599i+483=307+599i
(-16+21i)(11-23i) = -176 + 368i + 231i - 483i^2
= -176 + 599i + 483 = 307 + 599i

=307+599i442=307442+599442i
= \frac{307 + 599i}{442} = \frac{307}{442} + \frac{599}{442}i


Answer: 307442+599442i\dfrac{307}{442} + \dfrac{599}{442}i.
4If x−iy=a−ibc−idx - iy = \sqrt{\dfrac{a-ib}{c-id}}, prove that (x2+y2)2=a2+b2c2+d2(x^2+y^2)^2 = \dfrac{a^2+b^2}{c^2+d^2}.Show solution
Given: x−iy=a−ibc−idx - iy = \sqrt{\dfrac{a-ib}{c-id}}

Step 1: Square both sides:
(x−iy)2=a−ibc−id
(x-iy)^2 = \frac{a-ib}{c-id}

x2−y2−2xyi=a−ibc−id
x^2 - y^2 - 2xyi = \frac{a-ib}{c-id}


Step 2: Take modulus of both sides. Recall ∣z2∣=∣z∣2|z^2| = |z|^2 and ∣z1z2∣=∣z1∣∣z2∣\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}:
∣(x−iy)2∣=∣a−ibc−id∣
|(x-iy)^2| = \left|\frac{a-ib}{c-id}\right|

∣x−iy∣2=∣a−ib∣∣c−id∣
|x-iy|^2 = \frac{|a-ib|}{|c-id|}

(x2+y2)=a2+b2c2+d2
(x^2+y^2) = \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}


Step 3: Square both sides:
(x2+y2)2=a2+b2c2+d2
(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}


Hence proved.
5If z1=2−i, z2=1+iz_1 = 2-i,\ z_2 = 1+i, find ∣z1+z2+1z1−z2+1∣\left|\dfrac{z_1+z_2+1}{z_1-z_2+1}\right|.Show solution
Given: z1=2−i, z2=1+iz_1 = 2-i,\ z_2 = 1+i

Step 1: Compute numerator z1+z2+1z_1+z_2+1:
(2−i)+(1+i)+1=4+0i=4
(2-i)+(1+i)+1 = 4+0i = 4


Step 2: Compute denominator z1−z2+1z_1-z_2+1:
(2−i)−(1+i)+1=2−i−1−i+1=2−2i
(2-i)-(1+i)+1 = 2-i-1-i+1 = 2-2i


Step 3: Compute the modulus:
∣42−2i∣=∣4∣∣2−2i∣=44+4=422=22=2
\left|\frac{4}{2-2i}\right| = \frac{|4|}{|2-2i|} = \frac{4}{\sqrt{4+4}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}


Answer: 2\sqrt{2}.
6If a+ib=(x+i)22x2+1a + ib = \dfrac{(x+i)^2}{2x^2+1}, prove that a2+b2=(x2+1)2(2x2+1)2a^2+b^2 = \dfrac{(x^2+1)^2}{(2x^2+1)^2}.Show solution
Given: a+ib=(x+i)22x2+1a + ib = \dfrac{(x+i)^2}{2x^2+1}

Step 1: Expand the numerator:
(x+i)2=x2+2xi+i2=x2−1+2xi
(x+i)^2 = x^2 + 2xi + i^2 = x^2 - 1 + 2xi

a+ib=x2−12x2+1+2x2x2+1i
a + ib = \frac{x^2-1}{2x^2+1} + \frac{2x}{2x^2+1}i


So a=x2−12x2+1a = \dfrac{x^2-1}{2x^2+1} and b=2x2x2+1b = \dfrac{2x}{2x^2+1}.

Step 2: Compute a2+b2a^2+b^2:
a2+b2=(x2−1)2+4x2(2x2+1)2=x4−2x2+1+4x2(2x2+1)2=x4+2x2+1(2x2+1)2=(x2+1)2(2x2+1)2
a^2+b^2 = \frac{(x^2-1)^2 + 4x^2}{(2x^2+1)^2} = \frac{x^4 - 2x^2 + 1 + 4x^2}{(2x^2+1)^2} = \frac{x^4+2x^2+1}{(2x^2+1)^2} = \frac{(x^2+1)^2}{(2x^2+1)^2}


Hence proved.
7Let z1=2−i, z2=−2+iz_1 = 2-i,\ z_2 = -2+i. Find (i) Re⁡(z1z2zˉ1)\operatorname{Re}\left(\dfrac{z_1 z_2}{\bar{z}_1}\right), (ii) Im⁡(1z1zˉ1)\operatorname{Im}\left(\dfrac{1}{z_1\bar{z}_1}\right).Show solution
Given: z1=2−i, z2=−2+iz_1 = 2-i,\ z_2 = -2+i

So zˉ1=2+i\bar{z}_1 = 2+i.

(i) Find Re⁡(z1z2zˉ1)\operatorname{Re}\left(\dfrac{z_1 z_2}{\bar{z}_1}\right):

Step 1: Compute z1z2z_1 z_2:
(2−i)(−2+i)=−4+2i+2i−i2=−4+4i+1=−3+4i
(2-i)(-2+i) = -4+2i+2i-i^2 = -4+4i+1 = -3+4i


Step 2: Divide by zˉ1=2+i\bar{z}_1 = 2+i:
−3+4i2+i=(−3+4i)(2−i)(2+i)(2−i)=−6+3i+8i−4i24+1=−6+11i+45=−2+11i5
\frac{-3+4i}{2+i} = \frac{(-3+4i)(2-i)}{(2+i)(2-i)} = \frac{-6+3i+8i-4i^2}{4+1} = \frac{-6+11i+4}{5} = \frac{-2+11i}{5}


Re⁡(z1z2zˉ1)=−25
\operatorname{Re}\left(\frac{z_1 z_2}{\bar{z}_1}\right) = -\frac{2}{5}


(ii) Find Im⁡(1z1zˉ1)\operatorname{Im}\left(\dfrac{1}{z_1\bar{z}_1}\right):

Step 1: z1zˉ1=∣z1∣2=(2)2+(−1)2=5z_1\bar{z}_1 = |z_1|^2 = (2)^2+(-1)^2 = 5 (a real number).

Step 2: 1z1zˉ1=15\dfrac{1}{z_1\bar{z}_1} = \dfrac{1}{5}, which is purely real.

Im⁡(1z1zˉ1)=0
\operatorname{Im}\left(\frac{1}{z_1\bar{z}_1}\right) = 0


Answers: (i) −25-\dfrac{2}{5}, (ii) 00.
8Find the real numbers xx and yy if (x−iy)(3+5i)(x-iy)(3+5i) is the conjugate of −6−24i-6-24i.Show solution
Given: (x−iy)(3+5i)=(−6−24i)‾=−6+24i(x-iy)(3+5i) = \overline{(-6-24i)} = -6+24i

Step 1: Expand the left side:
(x−iy)(3+5i)=3x+5xi−3yi−5yi2=3x+5y+(5x−3y)i
(x-iy)(3+5i) = 3x + 5xi - 3yi - 5yi^2 = 3x + 5y + (5x-3y)i


Step 2: Equate real and imaginary parts:
3x+5y=−6⋯(1)
3x + 5y = -6 \quad \cdots (1)

5x−3y=24⋯(2)
5x - 3y = 24 \quad \cdots (2)


Step 3: Solve the system:

Multiply (1) by 3 and (2) by 5:
9x+15y=−18
9x + 15y = -18

25x−15y=120
25x - 15y = 120

Adding: 34x=102⇒x=334x = 102 \Rightarrow x = 3

Substitute in (1): 9+5y=−6⇒y=−39 + 5y = -6 \Rightarrow y = -3

Answer: x=3, y=−3x = 3,\ y = -3.
9Find the modulus of 1+i1−i−1−i1+i\dfrac{1+i}{1-i} - \dfrac{1-i}{1+i}.Show solution
Step 1: Simplify 1+i1−i\dfrac{1+i}{1-i}:
1+i1−i=(1+i)2(1−i)(1+i)=1+2i−12=2i2=i
\frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{1+2i-1}{2} = \frac{2i}{2} = i


Step 2: Simplify 1−i1+i\dfrac{1-i}{1+i}:
1−i1+i=(1−i)2(1+i)(1−i)=1−2i−12=−2i2=−i
\frac{1-i}{1+i} = \frac{(1-i)^2}{(1+i)(1-i)} = \frac{1-2i-1}{2} = \frac{-2i}{2} = -i


Step 3: Subtract:
i−(−i)=2i
i - (-i) = 2i


Step 4: Find modulus:
∣2i∣=2
|2i| = 2


Answer: 22.
10If (x+iy)3=u+iv(x+iy)^3 = u+iv, then show that ux+vy=4(x2−y2)\dfrac{u}{x} + \dfrac{v}{y} = 4(x^2-y^2).Show solution
Given: (x+iy)3=u+iv(x+iy)^3 = u+iv

Step 1: Expand (x+iy)3(x+iy)^3 using (a+b)3=a3+3a2b+3ab2+b3(a+b)^3 = a^3+3a^2b+3ab^2+b^3:
(x+iy)3=x3+3x2(iy)+3x(iy)2+(iy)3
(x+iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3

=x3+3x2yi+3x(−y2)+(−iy3)
= x^3 + 3x^2 yi + 3x(-y^2) + (-iy^3)

=(x3−3xy2)+i(3x2y−y3)
= (x^3 - 3xy^2) + i(3x^2y - y^3)


Step 2: Equate real and imaginary parts:
u=x3−3xy2,v=3x2y−y3
u = x^3 - 3xy^2, \quad v = 3x^2y - y^3


Step 3: Compute ux+vy\dfrac{u}{x} + \dfrac{v}{y}:
ux=x3−3xy2x=x2−3y2
\frac{u}{x} = \frac{x^3-3xy^2}{x} = x^2 - 3y^2

vy=3x2y−y3y=3x2−y2
\frac{v}{y} = \frac{3x^2y - y^3}{y} = 3x^2 - y^2

ux+vy=x2−3y2+3x2−y2=4x2−4y2=4(x2−y2)
\frac{u}{x} + \frac{v}{y} = x^2 - 3y^2 + 3x^2 - y^2 = 4x^2 - 4y^2 = 4(x^2-y^2)


Hence proved.
11If α\alpha and β\beta are different complex numbers with ∣β∣=1|\beta|=1, then find ∣β−α1−αˉβ∣\left|\dfrac{\beta-\alpha}{1-\bar{\alpha}\beta}\right|.Show solution
Given: ∣β∣=1|\beta|=1, α≠β\alpha \neq \beta.

Step 1: Compute ∣β−α1−αˉβ∣2\left|\dfrac{\beta-\alpha}{1-\bar{\alpha}\beta}\right|^2:
∣β−α1−αˉβ∣2=∣β−α∣2∣1−αˉβ∣2
\left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right|^2 = \frac{|\beta-\alpha|^2}{|1-\bar{\alpha}\beta|^2}


Step 2: Expand ∣β−α∣2=(β−α)(β−α)‾=(β−α)(βˉ−αˉ)|\beta-\alpha|^2 = (\beta-\alpha)\overline{(\beta-\alpha)} = (\beta-\alpha)(\bar{\beta}-\bar{\alpha})

Step 3: Expand ∣1−αˉβ∣2=(1−αˉβ)(1−αˉβ)‾=(1−αˉβ)(1−αβˉ)|1-\bar{\alpha}\beta|^2 = (1-\bar{\alpha}\beta)\overline{(1-\bar{\alpha}\beta)} = (1-\bar{\alpha}\beta)(1-\alpha\bar{\beta})

Step 4: Expand both:
∣β−α∣2=∣β∣2−βαˉ−βˉα+∣α∣2=1−βαˉ−βˉα+∣α∣2
|\beta-\alpha|^2 = |\beta|^2 - \beta\bar{\alpha} - \bar{\beta}\alpha + |\alpha|^2 = 1 - \beta\bar{\alpha} - \bar{\beta}\alpha + |\alpha|^2

∣1−αˉβ∣2=1−αˉβ−αβˉ+∣α∣2∣β∣2=1−αˉβ−αβˉ+∣α∣2
|1-\bar{\alpha}\beta|^2 = 1 - \bar{\alpha}\beta - \alpha\bar{\beta} + |\alpha|^2|\beta|^2 = 1 - \bar{\alpha}\beta - \alpha\bar{\beta} + |\alpha|^2


Since ∣β∣2=1|\beta|^2 = 1, both expressions are equal.

∣β−α1−αˉβ∣2=1  ⟹  ∣β−α1−αˉβ∣=1
\left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right|^2 = 1 \implies \left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right| = 1


Answer: 11.
12Find the number of non-zero integral solutions of the equation ∣1−i∣x=2x|1-i|^x = 2^x.Show solution
Given: ∣1−i∣x=2x|1-i|^x = 2^x

Step 1: Compute ∣1−i∣|1-i|:
∣1−i∣=12+(−1)2=2
|1-i| = \sqrt{1^2+(-1)^2} = \sqrt{2}


Step 2: Substitute:
(2)x=2x
(\sqrt{2})^x = 2^x

2x/2=2x
2^{x/2} = 2^x


Step 3: Equate exponents:
x2=x  ⟹  x=2x  ⟹  x=0
\frac{x}{2} = x \implies x = 2x \implies x = 0


The only solution is x=0x = 0, which is not a non-zero integer.

Answer: There is no non-zero integral solution (the number of non-zero integral solutions is 00).
13If (a+ib)(c+id)(e+if)(g+ih)=A+iB(a+ib)(c+id)(e+if)(g+ih) = A+iB, then show that (a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2+B^2.Show solution
Given: (a+ib)(c+id)(e+if)(g+ih)=A+iB(a+ib)(c+id)(e+if)(g+ih) = A+iB

Step 1: Take modulus of both sides:
∣(a+ib)(c+id)(e+if)(g+ih)∣=∣A+iB∣
|(a+ib)(c+id)(e+if)(g+ih)| = |A+iB|


Step 2: Use the property ∣z1z2∣=∣z1∣∣z2∣|z_1 z_2| = |z_1||z_2|:
∣a+ib∣⋅∣c+id∣⋅∣e+if∣⋅∣g+ih∣=∣A+iB∣
|a+ib|\cdot|c+id|\cdot|e+if|\cdot|g+ih| = |A+iB|

a2+b2⋅c2+d2⋅e2+f2⋅g2+h2=A2+B2
\sqrt{a^2+b^2}\cdot\sqrt{c^2+d^2}\cdot\sqrt{e^2+f^2}\cdot\sqrt{g^2+h^2} = \sqrt{A^2+B^2}


Step 3: Square both sides:
(a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2
(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2+B^2


Hence proved.
14If (1+i1−i)m=1\left(\dfrac{1+i}{1-i}\right)^m = 1, then find the least positive integral value of mm.Show solution
Step 1: Simplify 1+i1−i\dfrac{1+i}{1-i}:
1+i1−i=(1+i)(1+i)(1−i)(1+i)=(1+i)21+1=1+2i−12=2i2=i
\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{(1+i)^2}{1+1} = \frac{1+2i-1}{2} = \frac{2i}{2} = i


Step 2: The equation becomes:
im=1
i^m = 1


Step 3: Since i4=1i^4 = 1 and the powers of ii cycle with period 4, the smallest positive integer mm for which im=1i^m = 1 is:
m=4
m = 4


Answer: The least positive integral value of mm is 4\boxed{4}.

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