Complex Numbers Class 11
Master Complex Numbers Class 11 with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.
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Complex Numbers Class 11 – NCERT Solutions
Each question below opens its complete step-by-step Teachoo solution.
Ex 4.1
14 questionsEx 4.1, 1
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Ex 4.1,1
Express the given Complex number in the form a + ib:
. 3.
co (2)
. 3,
(si) (-5 4)
=5x = x (ix i)
=-3xixi
=-3x i?
=-3x-1 (Putting i? = -1)
=3
=3+0
=3+i0
Ex 4.1, 2
Ex 4.1,2 teackoo.com
Express the given Complex number in the form a + ib: i° + i19
(24729
=ix i8+jx {8
=ix (2)44+ix (2)?
Putting ? =—-1
=ix (-1)*+i~ (-1)°
=ix(1)+ix(-1)
=i-i
=0
=0+i0
Ex 4.1, 3
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Ex 4.1, 3
Express the following in the form a+ ib : i-°?
j-39 1
t ~ 439
ae!
~ G38) xi
_ 1
~ G2) xt
Putting i?=-—1
_ 1
~ C1) xi
_ 1
~ =41xi
_ 1
~ Hi
Ex 4.1, 4
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Ex 4.1, 4
Express the given Complex number in the form a + ib:
3(7 +77) +i(7 +77)
3(7+17)+i(7+i7)
=3x7+i7x3+ix7+ixi7
=214+211+71+i77
Putting ? = -1
=21+21i1+7i+(-1)7
=21+21i+7i-7
=21-7+21i+7i
=(21-7)+(217+77)
=14+28i
Ex 4.1, 5
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Ex 4.1, 5
Express the given Complex number in the form a + ib:
(1-1) — (-1 + i6)
(1-7) -(-1+76)
=(1-/) -(-1)-i6
=1-i+1-i6
= 1+1-i-f6
= (1+ 1)+-1-6)
=2+(-7i)
=2-7i
Ex 4.1, 6
Express the given Complex number in the forma + ib:
View solutionEx 4.1, 7
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Ex 4.1, 7
Express the given Complex number in the form a + ib:
1 7 a 4 >
(G+iz)+(4+i3)-(-3 +8)
1,.7 .1 4,
(G+iz)+ (4415) -(-$+9)
3 3 3 3
=t+i724a44it44-i
3 3 3.3
=(F+4+5)+i€ +2 -1)
3 3 3 3
(eet) j (8)
= (——— } +7 [| ————_
3 3
(2) (4)
= (——— _} + 7 | ———_
3 3
17 [5
-(F)+#G)
3 3
Ex 4.1, 8
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Ex 4.1, 8 (Method 1)
Express the given Complex number in the form a+ ib: (1- i)*
(1- i
=(a- oy
=(1-i)?(- i)?
Using (a -b)? =a? +b?-2ab
=(174+i7?-2x1xij)(1?4+i?-2x1x i)
=(14 i7-21)(1 4+ i?-2i)
=(1—-1-2i)(1-1-2i) (Putting i? =-1)
= (0- 21) (0- 21)
=(-2i)(-2i)
=4i?
Ex 4.1, 9
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Ex 4.1, 9
1 3
Express the given Complex number in the form a + ib: G + 3i)
3
1 .
(G+ 3i)
Itis the form (a+ bP
(a+ bP =a? +b’ + 3ab (a+b)
Putting a == and b=3i
3
-(i ;)3 tya;(2 i
=(2) +(39+3 xx 3i($+3:)
-i 3¢n34a7(1 ;
= it) () +3i($+ 3i)
_1 3,9:{1 .
=5,¢27i +3i(5+ 3i)
=24271343ix243ix3i
27 3
Ex 4.1, 10
Ex4. 1, 10 teachoo.com
Express the given Complex number in the form a + ib:
3
1.
(-2-31)
3
1,
(-2-51)
3
1,
3
1,
= - (2439)
It is of the form (a+b)?
Using (a+b)? =a? +b? +3 ab (a+b)
Here a=2and b=~i
3
__ 341, 1, 1,
= (@ +(5i) +3 x2xii(2+4))
Ex 4.1, 11
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Ex 4.1, 11
Find the multiplicative inverse of the Complex number 4 - 3i
Multiplicative inverse of z=z~+
Multiplicative inverse of z =<
Putting z= 4-3i
multiplicative inverse of 4-3i = =
Rationalizing
1 x 443i 443i
“4-31 443i (4-304 + 3)
Using (a-b)(a+b)=a?-b?
_ 443i
~ (4°-Gi?
_ 443i
~ 16-97
Ex 4.1, 12
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Ex 4.1, 12
Find the multiplicative inverse of the Complex number V5 + 3i
Multiplicative inverse of z =z~1
Multiplicative inverse of z = :
Putting z= V5 + 3i
multiplicative inverse of V5 + 3i = Faas
Rationalizing
-i,. V5 - 3i
~¥543i ~~ ¥5-3i
_ V5 — 3i
~ (V5 + 31)(V5- 31)
Using {a — b) {a + b) = a? - b?
___v¥5-3i
mC imxcon
_ V5 —3i
~ 5-97
Ex 4.1, 13
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Ex 4.1, 13
Find the multiplicative inverse of the Complex number - i
Multiplicative inverse of z =z~+
Multiplicative inverse of z ==
Putting z=-i
multiplicative inverse of —i =F
Multiplying and dividing by i
1 vi
=TXs
—t t
_i
“ip
Putting i? = -1
oi
--)
si
Ex 4.1, 14
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Ex 4.1, 14
Express the following Expression in the form of a + ib.
(3 + ivs } (3 - ivs)
(3 + V2) - WE -N2)
(3 +ivs ) (3-iv5)
(342i) - (3-12)
Using (a + b) (a - b) = a? - b?
_ _@P=(ivs)’
~ ¥34¥2i-v3+iv2
_ 9-i?x5
~ VB — V3 + v2i t iv2
Putting i? =—-1
_ 9-C1)Xx5
~ O4V2i+ iv2
_ 9+5
~ Zit y2i
_ 14
~ 2y2i
Examples
19 questionsExample 1
Exampl el teachoo.com
If 4x + i(3x — y) = 3 + i (-6), where x and y are real numbers, then find
the values of x and y.
Given
4x + i(x-y) =3 + i(-6)
Comparing Real Part Comparing Imaginary Part
4x = 3 3x-y = -6
Putting x ==
x=3 utting x =7
4
3
3(G)-y =-6
9
4 y=~6
9
4 +6= y
33
ya
Example 2 (i)
Example 2 teachoo.com
Express the following in the form of a + ib:
(i) (51) (Fé)
«i fl. 1 oe
(-5i) (i) =-5 x= xixi
= S xi? (Putting i = V¥—1)
= = x (v-1 2
-5
=a (-1)
==
“8
= 5 +0
8
=2 + i0
8
Example 2 (ii)
Example 2
Express the following in the form of a + bi
(ii) ( ) (2 ) ( 1/8 )^3
Example 3
Example 3 feachoo-tom
Express (5 — 3i)° in the form a + ib.
(5-3i)3
Using (a — b}? = a? - b?- 3ab (a—b)
Putting a = 5, b = 3i
= 53 — (3/)?-3 x5 x (3/) (5-3)
= 125 - 27/3 45i (5 -3/)
= 125 - 27 ix (?)-45ix (5) +457 x (3)
= 125 - 27 i (72)- 2251+ 135(?)
As ? =-1
= 125 - 27 i(-1)- 225 / + 135(-1)
=1254+27/-225/-135
Example 4
Example 4 feaclloo.com
Express (—V3 + V—2 ) (2V3 — i) in the form of a + ib
(-v3 + V=2 ) (2v3 - a)
= (-v3+V-1 x 2) (2v3-i)
= (-v3+V-1 x v2 }(2v3-i)
Putting V— 1 =i
= (-v3+iv2) (2 V3 —i)
= (-v3) (2 v3 -i)+ v2 (2 V3-1)
= —¥3 x (2V3 )- v3 x (i) + iV2 x (2V3) +iV2x(-d
=-3X2+V3i+2iW2V3-i2v2
Example 5
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Example 5
Find the multiplicative inverse of 2 — 3i.
Multiplicative inverse of 2 — 3i = oa
Rationalizing
1 24+3i
= nan * G+st)
(2-3) (243i)
7 24+3i
~ (2-3) (2 +38)
Using (a + b) (a-—b) = a? - b?
243i
~ 22 — (38)2
_ 243%
4-992
Example 6 (i)
Example 6 teachoo.com
Express the following in the form a + ib
. 5+¥2i
i) a
542i
1-¥2i
Rationalizing
_(Stv2i) (itv2i)
© (a-vai) ~ (2 +v2i)
* (Vai) (1+ V2)
~ (1)? — (Vai)? (Using a?- b? =(a + b) (a—b))
~ 1 -2i?
Example 6 (ii)
Example 6
Express the following in the form a+ ib
(ii) ^( 35)
Example 7
Example 7 teachoo.com
. . (3 — 2i)(2 + 3i)
Find the conjugate of (G@42)@=1)
First we calculate @—2DG+ 3) |
(14+ 21)(2-a
And then find its conjugate
Now,
(3 — 2i)(2 + 3i)
(44 2i)(2 — i)
_ 3(2 + 3i) — 2i(2 + 3i)
~ 1(2-i) + 2i(2 -i)
_3xX243x 3i- 21x 2-2ix 3i
~ Q-i42ix2-2ixi
_ 6+ 9i- 4i + 67?
~ 2-14 41-27?
Example 8
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Example 8
. _ (a+ ib)
Ilfx+iy = (a~ib)
Taking R.H.S
atib
a-ib
Rationalizing
__atib (atib)
~~ a-ib (a+ib)
_ _ (atib)?
~~ (a-ib)(a+ib)
_ a24+(ib)24+2aib (a+ by? =a7b? + 2ab
a?—(ib)? (a+ b)(a— b) = a?b?
Question 1
a
View solutionQuestion 2
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Example 8
16.
Convert the complex number Tei into polar form.
16
Let z= 14
Rationalizing
= 716, 1-iv3
“4a4iv3 1-iv3
= 7161-13) _
~ (1 +iv3) (4 -i¥3)
Using (a—b}(a+b)=a?-b?
_ —16(1-iv3)
~ OY =G3P
_-16(1-iv3)
~ 1-37
Question 3
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Example 9
Solve x? + 2 = 0
wr+2=0
xt= 0-2
xe =-2
x =t¥-2
=tV-1x2
=ty—-1xV2
=tixy2 fizV-T)
=4y2i
Question 4
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Example 10
Solvex? +x+1=0
x +x+1=0
The above equation is of the form
ax?+bxtc=0
Wherea=1,b=1,c=1
—-b+V b? -4ac
Here, x = ——————_
2a
Putting values of a, bandc
-14v1 -—4x1x1
x =
2x1
_ -1tvi-4
~ 2
_ -1ltiv=3
~ 2
Question 5
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Example, 11
Solve ¥5x2 +x + V5 =0
V5x24+x+V5=0
The above equation is of the form
axe + bx +c=0
where a=V5,b=1,andc=¥V5
_—b+Vv b? —4ac
x= 2a
Putting values
_-1ttV¥2?-4xV5xV5
x= 2x5
_-14v1-4Xx5
~ 2v5
_ -14+v1=20
~ 2V5
Question 6 (i)
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Example, 13
Find the modulus and argument of the complex numbers:
«iti
(i) Toi’
First we solve iit
1-t
Let Z= at
1-i
Rationalizing the same
1+i | 1ti
= — x —
1-i 1t+i
_ (4+i) +i)
~ (1-4) Gti)
Using (a — b) (a + b) = a? - b?
_ (1+iy’
(1ay-Ciy
Using (a +b)? = a2+ b? + 2ab
_ G)?+ + 2i
(1)? - @?
Question 6 (ii)
Example, 13
Find the modulus and argument of the complex numbers:
(ii) 1/(1 + 𝑖)
Question 7
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Example, 15
. 3+2isindO .
Find real 8 such that -—~—— is purely real
1 -2isin 8
Since 342i sin@ is purely real
1 - 2isin 6 P y
. 34 2isin®
We need to first solve —-———_ and then take
1-2isin®
imaginary part as 0
3 4+ 2isin®
1 - 2isin 8
Rationalizing
_3+2isin® x 1+ 2isin 8
~ 1-2isin@ © 1+ 2isin®
_ @+2isin 6) (1+ 2i sin 0)
~ (1 2isin 6)(1 + 2i sin 6)
_ 3(1+2isin ®) + 2isin 6 (1+ 2isin®)
~ (1 —2isin 6)(4 + 2isin @)
Question 8
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Example 16
i-1 :
Convert the complex number z = —-+~———- in the polar form.
cos 3 +isin z
i-1
Let z=——z——__—a
cos ~+isin —
3 3
_ i-1
cos( #2") +isin (=)
3 3
_ i-1
~ cos 60° + i sin 60°
a!
“4 ¥3,
z + at
_i-l
~ 4 + VBE
2
_2(i-1)
~ 143i
Rationalizing
_2Ci=1) y 1-v3i
~ 443i 1-V3i
Miscellaneous
21 questionsMisc 1
teachoo.com
Misc 1
25\3
. f 18 1
Evaluate: (i + (5) )
25,3
-18 1
(+) )
3
— {718 4 1
= (i + oe)
3
- (;18 1
~ (i Tix )
3
.2\9 1
= I + ——
( ) =isn)
Putting i? = —1
3
={¢—1? 1
-(¢ 1) tote)
Misc 2
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Misc 2
For any two complex numbers z, and 2,, prove that
Re (2,22) = Re z, Re Zz - Imz,Imz,
Complex number is of form
z=xt+tly
Hence
Let complex number z, = x, + ly,
Let complex number z, = Xz + ly,
Solving RHS first
Rez, Rez, — Imz, Imz,
aX, X2 —Viy2
Misc 3
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Misc 3
2 3-41
Reduce (=< - =) (=) to the standard form.
1+ 4i 1t+i S+i
We have to solve and then make it in Standard form a + ib
( 1 2 \(*)
144i 1+i/ \5+4i
_ (a2 —201— 40) (*)
TX a-49040 5+i
= (28) C*)
“Aasi-a4i-a?] Vsti
(Putting i? =-1)
=( 1+i-24+8i \(E*)
“Na 4i-4i-4ax(-y/ \sti
= (tnt) C*)
“Nasi-aita] \s+i
2 (Gaasi8i) (a-#)
“\a444i-4i 5+i
Misc 4
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Misc 4 (Introduction)
. ‘a— ib _ a? + b?
Ifx-iy= |-—— prove that (x? + y*)? = aaa
(x- iy) @+ iy)
Using (a-b)(a+b)=a?-b?
= (x)? - (Gy)
=x? (Dy?
=x?- (—1);” (As i2=— 1)
= x2 + y
Misc 5
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Misc 5 (Method 1)
. _ |4atz24¢1
If2,=2-i,2)= 141, find [272 "|
Zy-Zgt1
z=xtiy
. Zy+Z24+1
We have to find ee Modulus of z = [z/
Zy—Zz4+1
= [z+ yy?
. . Zy+Z24+1
First we find ~—2—
Zy—Z24+1
zt+Z.+1_(@—i)+(M+i)+1
Z-Z+1 (2-i)-(1+i)+1
_2-ititita
~2-i-1-i41
_24+1+1-iti
~2-141-i-1
Misc 6
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Misc 6 (Method 1)
_ _ (x+i? 2 2 _ (x?+1)
Ifat ib=s aa? prove that a? + b? = tata?
, +i)?
a+ ib = at
2x241
Using (a + b)? = a? + b? + Zab
_ 4)? + 2xi
~ 2x24
Putting i2 = —-1
_ x14 2xi
ox?
x1 . 2x
= +i-;—
2x24+1 2x241
Misc 7
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Misc 7
Let z,=2-i,z,=-2+i .Find
(i) Re (=)
1
z=atib
Z = conjugate
We need to find Re (22) Z=a—ib
1
i 7172,
i.e Real part of ( z )
Lets first calculate (22)
1
z,=2-i
z=-2+i
z= 2+i
Misc 8
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Misc 8
Find the real numbers x and y if (x - iy) (3 + 5i) is the
conjugate of -6- 241. peatib
Z = conjugate
Conjugate of —6 — 24i
=-—6 4+ 24i (1)
Now it is given that
(x- iy) (3 + 5i) is conjugateof-—6 + 24: --(2)
Hence from (1) and (2)
-—6 + 241 = (x-iy) 345i
—6 + 241 =x(34+5) - iy + 5i)
—6 + 241 = 3x4 5xi -— 3yi-5i?y
Putting i? = —-1
—64+24i = 3x + Sxi-—3yi-5x-1xy
Misc 9
Misc 9 teachoo.com
Find the modulus of —+ — +—
1-i 14+i
i Att i-i z=xtiy
First we solve —— 1t+i Modulus of z = [z/
Hi int = (Pty
1-i Lt+i
7 (1-8 (14+ i)
~ @-) G+)
Using a?- b? = (a + b} (a - b)
()? -@*
7 1-?
~ 1-2
Misc 10
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Misc 10
If (x + iy)? = u + iv, then show that - + 7 =4 (x? - y?).
We know that (a + b)3 = a3 + b3 + 3ab (a+ b)
Replacing a = x and b= iy
(x + iy)F= 23 + (iy) + 3x iy & t+ iy)
=x3 4 By3 + 3x yi (xt iy)
=8 4+ 2 xiyi+ 3x2yi + 38xy7i?
Putting i? =-1
=x34(-—1xi x xy”) + 3x? yit 3xy? x(-1)
=x3- iy? + 3x? yi — 3xy?
=x3- 3xy? — iy3 + 3x’yi
=x3- Bxy?+ 3x’yi — iy?
=x3- 3xy?24+ 32x2y -— yi
Misc 11
Misc 11 teachoo.com
If a and B are different complex numbers with || = 1, then
. B-aO
z=x+iy
We know that |z]? = (z) (Z} Z = conjugate
z=x-iy
—— |z| = modulus
B-a 2 _ B-a B-a
al (GD |z| = x2 + y?
_ B-a B-a
~ (ES) (5)
_ (8-« B-a
~ (ES) (3)
((@ =a@) & Conjugate of 1 is 1 i.e. 1 = 1)
_ B-a p-@
~ (ES) (55)
_ G-4)\(B-z)
(1-@B)(1-a@ B)
Misc 12
Misc 12 teachoo.com
Find the number of non-zero integral solutions of the equation
j1-i]* = 2%.
We need to find the value of x which should be an integer but
not 0
Lets first find the value of |1- i| zextly
Modulus of z = [z/
= f/x + y2
1-i y
Complex number is of the form x + iy
Where x=1
y=-1
[1 - i= x? $y?
=(@?+ C1?
=vIF1
=V2
Misc 13
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Misc 13 Introduction
If (a + ib)(c +id)}(e + if}(g + th) = A + iB, then show that
(a? + b?) (c? + d?) (e? + f2) (g2 +h?) = A? + B?.
(A + iB) (A -iB)
Using (a-b)(at+b)=a?-b
= A?- (iB)?
= A2- j2B2
Putting i2=—-1
= A*- (-1) B?
= A? +B
Hence, (A + iB) (A - iB) = A? + B?
Misc 14
Misc 14 teachoo.com
1+i\™ . rae
If (=) = 1, then find the least positive integral value of m.
We need to find minimum value of m which is positive as well as
integer.
Lets first find the value of (=)
1+i
1-i
Rationalizing
1+i 1+ti
= — xX —
1-i so 1ti
_ @+datd
~a-p)0+d
_ atriy F = @2—b2
= Gy we (Using (a + b) (a- b) = a? - b?)
Question 1 (i)
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Misc 5
Convert the following in the polar form:
4 L+7i
ory
147i
Let z= e-p
Using (a — b}? = a? + b? -— 2ab
_ (1+ 7i)
~ QyP+ GP -2x2xi
_ 147i
4t Poi
Putting i? = -1
_ 147i
“44 (-1)-4i
ati
“4-1-4i
_ ati
“3-41
Question 1 (ii)
Misc 5
Convert the following in the polar form:
(ii) ( 1 + 3𝑖)/(1 − 2𝑖)
Question 2
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Misc 6
Solve the equation 3x?- 4x + a =0
3x? - 4x + a =0
Multiplying both sides by 3
3 x (3x2 - 4x + *) =3x0
3x 3x?-3x Ax4+3x—" =0
9x?- 12x + 20 = 0
The above equation is of the form
ax? + bx +ce=0
Where a=9, b=-12, andc=20
_ —bAV b? -4ac
x= 2a
_ —(@12) 4 ¥(-12)? -4x 9x 20
~ 2x9
Question 3
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Misc 7
. 3
Solve the equation x?- 2x + 7 0
3
x?- 2x+==0
2
Multiplying whole equation by 2
3
2x x?- 2x Wx + 2x5=- 0x 2
2x7- 4x +3 =0
The above equation is of the form
ax? + bx +c¢=0
Where a=2,b=-4,andc=3
x= —b +V b? - 4ac
~ 2a
x= -(-4) 4+ ¥(-4)? -4x 2x3
~ 2x2
_ 4416-24
4
Question 4
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Misc 8
Solve the equation 27x?- 10x + 1 = 0
27x2- 10x +1=0
The above equation is of the form
axe+ bx +c=0
Where a = 27, b=-10, andc=1
_—b+vb? - 4ac
x= 2a
_-(-10) + f(@ 10)? = 4x (27) x G1)
~ 2x27
_ 10+ ¥100 = 108
~ 2x27
_ 10 + ¥-8
~ 2x27
_ 10 +¥=2 x4
~ 2x27
Question 5
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Misc 9
Solve the equation 21x? - 28x + 10 = 0
21x?- 28x +10 = 0
The Above equation of the form
ax? + bx +c =0
Where a=21,b=-28,andc=10
x = —b+v b? -4ac
~ 2a
x= (= 28) + ¥(-28)? -4 x 21 x10
~ 2x21
_ 284784 — 840
~ 42
_ 284V=56
~ 42
_ 284+ V=14x4
42
Question 6
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Misc 13
: 142i
Find the modulus and argument of the complex number 7 :
: 142i
First we solve ——
1-31
142i
Let z=——
1-3i
Rationalizing the same
_ 142i x it 3i
“41-31 143i
_ +2) (14+3)
~ (1-38) (1438)
104431) + 2°44 30
~ 4-3) 4+30
Modulus, Argument, Polar Representation
8 questionsQuestion 1
Ex 5.2, 1 teachoo
Find the modulus and the argument of the complex number
z=-1-iv3
(1)
Given z= —1-—iv3
—_ (2)
Letz=r(cos0+i sin@)
Here, ris modulus, and @ is argument
Comparing (1) & (2)
—1-ivV3=r(cos@+i sin0)
—1-ivV3=rcos@+irsin@
Comparing real and imaginary parts
Question 2
feachoo.com
Ex5.2, 2
Find the modulus and the argument of the complex number
z= —-vV3+ti -
Z=xtiy
Method (1) To calculate modulus of z cyxt ty?
z=-vV3+i
Complex number z is of the form x + iy
Where x =- V3 andy=1
Modulus of z
= |z]
= fx? 4 y2
- (yr +ay
=v341
=vi
=2
Question 3
teachoo.com
Ex5.2, 3
Convert the given complex number in polar form: 1 —i
Givenz = 1-i (1)
Let polar form be
z=r (cos 6 + isin 8) (2)
From (1) and (2)
1-i =r(cosO+isin 8}
4 4 4 J
Real Imaginary Real Imaginary
part part part part
Comparing real part Comparing imaginary part
1=rcos0 -1=rsin0
Squaring both sides Squaring both sides
(1)? = (r cos 6)? (— 1)? = (rsin6)?
1 = r*cos?8 (3) 1 = r’sin? 8 (4)
Question 4
teachoo.com
Ex 5.2, 4
Convert the given complex number in polar form: — 1+ i
Given z =-1+i (1)
Let polar form be
z= r(cos0+i sin®) (2)
From (1) & (2)
—1+i=r(cos@+i sin8)
—1+i =rcosO+tirsin®@
oy
Real Imaginary Real Imaginary
part part part part
Comparing Real part Comparing imaginary part
—1=r-cos@ 1 =rsin8
Squaring both sides Squaring both sides
(-1)2 = r? cos? 6 (1)? = (r sin)?
1= r*cos? 6 (3) 1 =r*sin26 wu (4)
Question 5
teackoo.com
Ex5.2, 5
Convert the given complex number in polar form: — 1—i
Given z=-1-i (1)
Let polar form be
z=r(cos6+isin8) (2)
From (1) & (2)
—1-i =r(cos0+i sin6)
-—1-i=rcosO+irsind
rr rr Sr
Real Imaginary Real Imaginary
part part part part
Comparing Real part Comparing imaginary part
—1=r-cos@ -1 =rsin@
Squaring both sides Squaring both sides
(-1)2 = r? cos? 6 (-1)? = (r sin)?
1= r*cos? 6 (3) 1 =r*sin26 wu (4)
Question 6
teachoo.com
Ex 5.2, 6
Convert the given complex number in polar form: -3
Givenz=—3
z=-3+0i (1)
Let polar form be
z=r{(cos@+isin 8) ...(2)
From (1) & (2)
—340f=r(cos6+isin 8}
—3+0i =rcosO+irsind
See ae
Real Imaginary Real Imaginary
part part part part
Comparing Real parts Comparing imaginary part
-3=rcos8 O=rsin@
Squaring both side Squaring both sides
(-3)?= (rcos 6}? {0)*=(rsin 6)?
9= 6 cos? (3) O= r’sin?6 (A)
Question 7
teachoo.com
Ex 5.2, 7
Convert the given complex number in polar form: V3 +i.
Given z= V3 +i (1)
Let polar form be
z=r(cos@+isin 9} (2)
From (1) & (2)
v3 +i=r(cos@ +isin9)
yf f i
Real Imaginary Real Imaginary
part part part part
Comparing Real part Comparing imaginary part
V¥3=rcos@ 1L=rsind
Squaring both sides Squaring both sides
(v3)? = (r cos 6) (1) =(r sin 0
3= rcos? 6 (3) 1= r’sin? 6 AA)
Question 8
teachoo.com
Ex 5.2, 8:
Convert the given complex number in polar form: i
zal
z=O +i (1)
Let polar form be z =r (cos 6 +i sin ®) (2)
From (1) & (2)
0+/=r(cos@+isin8)
0 +i =rcos@ + irsin®
a
Real Imaginary Real imaginary
part part part part
Comparing real part Comparing imaginary part
O=rcos8 1=rsin0
Squaring both sides Squaring both sides
(0)? = (rcos6)? (1)?= (rsin6
0=r? cos? 8 {3} 1=r’ sin? 6 (4)
Quadratic Equations with Complex Roots
10 questionsQuestion 1
Ex5.3, 1 teachoo
Solve the equation x? + 3 = 0
Now,
v?+3=0
x? = —3
x =47V-3
x =t+V¥-1 x3
x =+VJ-1x v3
x =+ixv3 (iz=V—-D
x =1v3i
Question 2
Ex 5.3, 2 feackoo.com
Solve the equation 2x* +x +1=0
2x2+x+1=0
The above equation is of the form
ax? + bx +c¢=0
Wherea=2,b=1,c=1
—b+v b? -4ac
x=
2a
_ rity)? -4x2x1
~ 2x2
_ -14+v1-8
7 4
Question 3
teackoo.com
Ex 5.3, 3
Solve the equation x? + 3x + 9 = 0
w+ 3x4+9=0
The above equation is of the form
ax? + bx +c=0
Where a= 1, b=3, andc=9
x= —b+V b? -4ac
~ 2a
-34V32-4x1x9
x=
2x1
_ -34 ¥9=36
~ 2
_ -3 4-27
~ 2
_ - 34-127
~ 2
Question 4
teackoo.com
Ex5.3, 4
Solve the equation -x? + x- 2 = 0
-x*7+x-2=0
The above equation is of the form
ax? + bx +c=0
Where a =-1, b=1, andc =-2
_—b+v b? -4ac
x= 2a
-14+ 7G)? -4x C1) (-2)
x =
2x(-1)
_ it f1=4% G2)
~ 2
_ 14v1-8
~ 32
_ -1tiv=7
~ -2
_ -itVv7xX=1
-2
Question 5
teackoo.com
Ex5.3,5
Solve the equation x? + 3x + 5 = 0
x?+3x4+5=0
The above equation is of the form
ax? + bx +c¢=0
Where a=1,b=3,c=5
x = DREN BPH Aac
~ 2a
34 ¥G)2-4x1x5
~ 2x1
_ -34+ ¥9=20
~ 2
_ -34¥=11
~ 2
_ -34V¥-1xT1
~ 2
_ -34y-1 x vit
~ 2
Question 6
teackoo.com
Ex5.3, 6
Solve the equation x*- x + 2 = 0
x¥-x+2=0
The above equation is of the form
ax? + bx +c = 0
Wherea=1,b=-1,c=2
yoke BF hac
~ 2a
xe -(-1) +1)? -4x1x2
~ 2x1
_ityi-8
~ 2
_ 1tVv-7
~ 2
_ ttv-1x7
~ 2
Question 7
teackoo.com
Ex5.3, 7
Solve the equation V2x? + x + ¥2 =0
V2x2 +x+¥2=0
The above equation is of the form
ax* + bx +¢ = 0
Where a=¥2,b=1,andc=V2
_ -bAvb? —4ac
x= 2a
14 JC)%-4x 2x v2
x= 2xV¥2
_ -ttvi=4x2
=a
_ -1+y1-8
=
Question 8
teachoo.com
Ex5.3, 8
Solve the equation ¥3x? - V2x + 3 ¥3 =0
V¥3x? - ¥2x+3V3=0
The above equation is of the form
ax* + bx +¢ = 0
Where a = V3 ,b=-V2, and c=3V3
_ bv b? - 4ac
x= 2a
~(-v2) + |(- v2)-4x v3 x 3v3
~ 2xV¥3
_v24+¥2—-4x3x3
SF
_ V2 4 ¥2—36
- SF
_ v2 + v= 34
-—
Question 9
teackoo.com
Ex5.3, 9
; 2 a
Solve the equation x* + Xt 0
2 tL
xe + xt 3 0
Multiplying both sides by V2
1
V2x2 + ¥2x + V2 x yr Ox v2
V2x2 +V2x+1=0
The above equation is of the form
ax? + bx +c¢=0
Where a=¥V2,b=V2, andc=1
x —b+vVb* —4ac
~ 2a
—v2 + | (v2) -4x 2x1
~ 2x2
Question 10
teackoo.com
Ex 5.3, 10
A 24% =
Solve the equation x’ +ytl 0
24 % =
x togtl 0)
Multiply the equation by V2
V2 x (x27 +54 1)=V2 x0
V2 xv + V2 xZ+V2 x1=0
V2x2+x+V¥2=0
The above equation is of the form
ax? + bx +c¢=0
Where a=v¥2,b=1,andc=v2
ye = DEN B? = Aae
~ 2a
_-1ty¥1?-4xy2xv2
~ 2x2
Why Learn This With Teachoo?
Complex Numbers introduces a number system in which equations such as x² + 1 = 0 have solutions. The chapter defines the imaginary unit i, develops algebra with numbers of the form a + ib, connects complex numbers with quadratic equations and introduces conjugate, modulus, argument and polar representation. Teachoo provides NCERT exercise solutions, examples, miscellaneous questions and concept-wise practice for Class 11 Complex Numbers, including complex roots of quadratic equations.
What are complex numbers?
The imaginary unit i is defined by i² = −1. A complex number is written as z = a + ib, where a and b are real. Here, a is the real part Re(z) and b is the imaginary part Im(z). A real number is a complex number with b = 0, while a purely imaginary number has a = 0.
Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal. Addition and subtraction combine corresponding parts. Multiplication uses ordinary algebra together with i² = −1. Division is simplified by multiplying numerator and denominator by the conjugate of the denominator.
The powers of i repeat in a cycle of four: i, −1, −i, 1. Therefore a large exponent can be reduced modulo 4. Students also apply algebraic identities such as (z₁ + z₂)² and factorisations while preserving the rule i² = −1.
Conjugate, modulus and Argand plane
If z = a + ib, its conjugate is z̄ = a − ib. Important results include z + z̄ = 2a, z − z̄ = 2ib and z z̄ = a² + b². The modulus is |z| = √(a² + b²), representing the distance of the point (a, b) from the origin in the Argand plane.
Geometrically, every complex number corresponds to a point or directed line segment in a plane whose horizontal axis is real and vertical axis is imaginary. The argument θ describes the angle made with the positive real axis. Quadrant information is essential when finding θ; using tan θ = b/a alone can produce the wrong angle.
The polar form is z = r(cos θ + i sin θ), where r = |z|. It reveals both magnitude and direction and prepares students for later results involving multiplication, division and powers of complex numbers.
Complex roots of quadratic equations
For ax² + bx + c = 0, the quadratic formula remains x = [−b ± √(b² − 4ac)]/(2a). If the discriminant is negative, its square root is expressed using i. For real coefficients, non-real roots occur as a conjugate pair. Students must simplify the radical carefully and write each answer in a + ib form.
The relationships between roots and coefficients still apply. If α and β are roots, α + β = −b/a and αβ = c/a. These relationships can verify complex roots or help construct a quadratic equation.
Topics covered on Teachoo
-
Exercise 4.1, NCERT examples and miscellaneous questions;
-
definition and algebra of complex numbers;
-
equality of two complex numbers;
-
conversion to a + ib form;
-
algebraic identities involving complex numbers;
-
division using conjugates;
-
cyclic powers of i;
-
conjugate and its properties;
-
modulus and argument;
-
polar representation;
-
quadratic equations with complex roots;
-
proof questions using conjugates, moduli or general complex numbers.
Key results to remember
-
i² = −1, i³ = −i and i⁴ = 1;
-
z + z̄ = 2 Re(z);
-
z − z̄ = 2i Im(z);
-
z z̄ = |z|²;
-
|z₁z₂| = |z₁||z₂|;
-
|z₁/z₂| = |z₁|/|z₂| for z₂ ≠ 0;
-
a complex number is zero only if both its real and imaginary parts are zero;
-
non-real roots of a real-coefficient quadratic occur in conjugate pairs.
Learning outcomes
Students should be able to simplify expressions involving i, perform all four arithmetic operations and compare complex numbers. They should use conjugates to rationalise a denominator, calculate modulus and argument, plot a complex number and convert between Cartesian and polar forms. They should solve quadratic equations with negative discriminants and prove standard properties using algebra.
Why is this chapter important?
Complex numbers extend algebra beyond the real line and make every quadratic equation solvable within a consistent system. They reappear in Class 12 and are important in JEE Mathematics, coordinate geometry, polynomial theory, electrical engineering and wave analysis. The Argand-plane viewpoint also strengthens the connection between algebra and geometry.
How Teachoo helps you prepare
Teachoo groups questions by concept, allowing students to master powers of i, division, conjugate, modulus, polar form and proof separately before attempting mixed problems. Serial-order solutions make it easy to locate a specific NCERT question, example or miscellaneous problem.
For every calculation, reduce powers of i early and collect real and imaginary terms at the end. In division, multiply by the conjugate of the complete denominator, not just one term. In argument questions, plot the point before selecting the angle. For quadratic roots, verify the final pair through sum and product whenever possible.
School-exam, JEE and competency preparation
School exams often ask for simplification, equality, roots, modulus and polar form. JEE questions can combine conjugates, loci, inequalities and quadratic-root relationships. Keep both algebraic and geometric meanings available: an equation involving |z − z₀| usually describes a distance from the point z₀.
Competency questions may model movement in a plane or ask students to diagnose a flawed calculation. Check whether the argument lies in the correct quadrant and whether the denominator was made real. In proof questions, start with z = a + ib when a direct property does not immediately apply.
Quick revision checklist
Simplify five large powers of i, perform one multiplication and one division, find the conjugate and modulus of three numbers, plot points in all four quadrants, convert one number to polar form and solve two quadratics with negative discriminants. Finish with a proof involving z and z̄.
Common mistakes to avoid
Do not write √(−a) = −√a; for positive a it is i√a. Do not treat i as an ordinary real variable or forget i² = −1 during expansion. The argument is not determined by tan⁻¹(b/a) without quadrant correction. Modulus is a non-negative real number, not a complex expression. Equality requires matching both parts.
Deeper reasoning and concept connections
A student has understood Complex Numbers only when the idea can be moved between words, diagrams, examples and mathematical notation. Start with a concrete example, identify what changes and what remains fixed, represent the relationship clearly and then state the rule. This movement between representations is important because school and competency questions often present a familiar idea in an unfamiliar form.
The chapter should also be connected to earlier and later mathematics. Definitions supply the language, worked examples reveal the method, and mixed questions test whether the method can be selected without a hint. Instead of memorising the appearance of a solved question, ask what information triggered the method, which condition made it valid and how the answer could be checked. That makes learning transferable to later chapters rather than limited to one exercise.
How to solve unfamiliar and competency-based questions
Read the complete problem before calculating. Underline the quantities, conditions and command word—find, compare, construct, justify, estimate or prove. Rephrase the task in one sentence and choose a representation such as a table, labelled figure, number line, expression or graph. Solve in small steps, keeping units and labels visible.
For an application question, the final line must answer the situation, not only display a number. For an assertion–reason question, test the assertion and reason separately before deciding whether one explains the other. For an MCQ, eliminate options using definitions, signs, size estimates or boundary cases before performing long calculations. If the answer is visual, check it against the stated scale or construction conditions rather than the appearance of the drawing.
What complete mastery looks like
For Complex Numbers, a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.
Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.
Additional frequently asked questions
What should a student know before starting Complex Numbers?
Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.
How can a student check an answer in Complex Numbers?
Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.
How many questions are enough for strong preparation?
There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.
How should Teachoo solutions be used without becoming dependent on them?
Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.
Frequently asked questions
Why do we need complex numbers?
They extend the real number system so polynomial equations such as x² + 1 = 0 can be solved consistently.
What is the conjugate of a complex number?
For z = a + ib, the conjugate is a − ib. Its product with z is the real number a² + b².
What is the modulus of z = a + ib?
It is √(a² + b²), the distance from (a, b) to the origin on the Argand plane.
Why do non-real quadratic roots occur in pairs?
For a polynomial with real coefficients, taking the conjugate of the equation shows that the conjugate of any non-real root is also a root.
Does Teachoo cover polar form and complex quadratic roots?
Yes. The chapter has focused sections for modulus, argument, polar representation and quadratic equations with complex roots, along with NCERT solutions.
Approach complex numbers as both algebra and geometry. The algebra handles calculation, while the Argand plane explains modulus, argument and many higher-level applications.