Skip to main content
Super TutorResources
Chapter 4 of 14
QuestionsNotesFlashcardsSummaryFormulasQuizConceptsSyllabusPlan
NCERT Solutions

Complex Numbers and Quadratic Equations

Meghalaya Board · Class 11 · Mathematics

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

101 questions20 flashcards5 concepts

Interactive on Super Tutor

Studying Complex Numbers and Quadratic Equations? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for ncert solutions and more.

Try Super Tutor — Free

1,000+ Class 11 students started this chapter today

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}.

Stuck on a step?

Ask Super Tutor AI to explain any solution on this page in a simpler way — free, 24x7.

Ask a Doubt Free

Frequently Asked Questions

What are the important topics in Complex Numbers and Quadratic Equations for Meghalaya Board Class 11 Mathematics?
Complex Numbers and Quadratic Equations covers several key topics that are frequently asked in Meghalaya Board Class 11 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Complex Numbers and Quadratic Equations — Meghalaya Board Class 11 Mathematics?
Understand the core concepts first, then work through the 101 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Complex Numbers and Quadratic Equations Class 11 Mathematics?
This page has free step-by-step NCERT Solutions for every exercise question in Complex Numbers and Quadratic Equations (Meghalaya Board Class 11 Mathematics) — written the way examiners award marks: given, formula, working, answer.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

More resources for Complex Numbers and Quadratic Equations

All 14 chapters — Meghalaya Board Class 11 Mathematics NCERT Solutions

For serious students

Get the full Complex Numbers and Quadratic Equations chapter — for free.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for Meghalaya Board Class 11 Mathematics.

Try Super Tutor Free
Try the full chapter — Free