Class 11 Chapter 5 Complex Numbers and Quadratic Equations (Polar form ) Lecture 3.pdf

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Class 11 Chapter 5 Complex Numbers and Quadratic Equations Lecture 3
#polar_form
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1. Dr. Pranav Sharma
Maths Learning Centre. Jalandhar.

2. POLAR REPRESENTATION OF COMPLEX NUMBERS

COMPLEX PLANE OR ARGAND PLANE

Let X'OX and YOY’ be the mutually perpendicular lines, known as the x-axis and
the y-axis respectively. The complex number (x + Ey) corresponds to the ordered pair
(x,y) and it can be represented by the point P(x, y) in the xy plane. The x-y plane
is known as the complex plane, or the Argand plane. x-axis is called the real axis
and y-axis is called the imaginary axis.

Note that every number on the X-axis is a real number, while each ow the y-axis is an
imaginary number.

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3. The complex numbers represented

geometrically in the diagram are Ml “iS

(3+50, (=4+ 30; *B(-4, 3)

(5-30, (2-44), 20.2)

(6+ 0i), (—3 + Où),

(0+ 2i) and (0 —i), x’ F(3,0) |° E(6, 0) X
represented by the points H(0, -1)

*C(-5, +3) Ga)
y

A(3,5), B(-4,3), €(-5, -3),D(2, —4), E(6, 0), F(—3,0), G(0,2) and H(0,—1)
respectively.

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4. POLAR FORM OF A COMPLEX NUMBER.

Let the complex number Z = x + iy be represented by the point P(x, y) in the complex
plane. Let ZXOP = 8 and |OP| =r > 0. Then,

P(r,0) are called the polar coordinates of P.

We call the origin O as pole.

x=rcos@andy=rsing

z=r(cos@+isin@). —————
This is called the polar form, ov trigonometric form,

or modulus-amplitude form, of Z.

Here, r = (x? + y? = |2| is called the modulus of Z.

And 6 is called the argument, or amplitude of 2,

written as arg (z), or amp(z) .

The value of O such that —1 < O < Wis called the principal argument of Z.

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5. METHOD FOR FINDING THE PRINCIPAL ARGUMENT OF A COMPLE
NUMBER
Case1 When z = (x + iy) lies on one of the axes:

L When z is purely real.

In this case Z lies on the x-axis.

(6) tf Z lies on positive side of the x-axis then O = 0.
(ti) tf z lies on negative side of the x-axis then O = 7.

I. When z is purely imaginary.
In this case, z Lies on the y-axis.
( tf z lies onthe y-axis and above the x-axis then 0 = 5.

(tt) tf z lies ow the y-axis and below the x-axis then 8 = ER

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6. Case 2 When g = (x+iy) aves not lie on any axes:

<
2 A Im (z) [e]
Step 1. Find the acute angle a given by tana = | a el 5
Step 2. Find the quadrant in which P(x,y) lies. >
Then, O = arg (Z) or amp(z) may be obtained as under. 8
Y Y 3
®
P

; a
4 Go|, > 2 E
Ki o M x x x ®
a
=.
wo u E
(1) When z lies in quad. I; (tt) When z lies in quad. II; then, O = =
then, 8 =a = arg(z) =a. (mw — a) > arg (z) = (r—a). ES
2
£
=
oO
un

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7. — (1 —a)

y UW

(tit) When z lies in quad. li
then, 8 =(a— 1) or (w+ a)
> arg(z) = (a—-m)or(r+a).

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(tv) When z lies in quad. IV; then, 0 =
—a or 0 = (2m — a)
> arg (z) = -a or (2m — à).

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8. Polar form of z = x+iy is r(cos O + isin 0).

<
() r= 12 =/x2+y?. (&) tana =/?| = ze. 2
When —Tr < 0 < Tr, then 8 is the principal argument of z. S
Quad. in which z Lies arg (z) 8
l 0=a 3
u e=n-a a
ut 0=-(n-a) <
IV 0=-aor (20 - a) 2
When z is purely real; then, z lies on the X-axis. ES
(x>0>0=0)aud(x<0>0=1). >
5
When z is purely imaginary; then, z lies on the y-axis. 9
(y>0>0=5) and (y<0>0= 2). 2
®
3
=
=
&

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9. Convert each of the following complex numbers into polar form:
( 3 (i)—5 (Go) i (iv) —2i

(t) The given complex number is Z = 3 + Oi.
Let its polar form be z = r( cos 0 + isin 0).
Now, r = |z| = (32 + 02 = V9 =3.
Clearly, Z = 3 + Oi is represented by the point P(3,0) , which Lies on the positive
side of the x-axis.
arg(z)=0=0.
Thus, r = 3 and 0 =0.
Hence, the required polar form of z = 3 + Oiiss(cos0+isin0).

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10. (ti) The given complex number is z = —5 + Oi.
Let its polar form be z = r( cos O + isin 0).
Now, r = |z| = (25)? + 02 = V25 = 5.
Clearly, Z = —5 + Oi is represented by the point P(—5, 0) , which lies on the
negative side of the x-axis.
arg(z)=1r>0=T.
Thus, r=50nd 8 = 1.
Hence, the required polar form of z = —5 + Oi ís 5 (cosr+isinr).

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11. (tii) The given complex numberis z = 0+ i.

Let its polar form be z = r( cos O + isin 0).

Now, r = |z| = V02 +12 = 1.

Clearly, z = 0 + dis represented by the point P(0, 1) , which Lies on the y-axis and
above the x-axis.

( ) TT 0 T
=3> ==,

. arg (Z)=5 2

This. r= Land 0 =>

Hence, the required polar form of Z = 0+ Lis

Big gop TA y Eo a ges E
1. (cos = +isin =) te, (cos 2 +isin =).

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12. (tv) The given complex number is z = 0 — Zi.

Let its polar form be z = r( cos O + isin 0).

Now, r = |z| = (0? + (—2)? = V4 = 2.

Clearly, z = (0 — 21) ts represented by the point P(0,—2), which lies on the y-axis
and below the x-axis.

( ) rn 8 TT
=— 0 -——

u arg (z 2 5

Thus, r= 2 and 0 ==.

Hence, the required polar form of z = 0 — 21 ís

z=2 { cos E +isin E Le., 2(cos E isin 3:

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13. Find the modulus and argument of each of the complex numbers given below:
(91+ (ii) — V3 +i (49 -1-iV3

© Leez=14 i. Then, |z| = 12 + 12 = V2.

Let à be the acute angle given by

Im (z)
Re (z)

ike 1 Tw
=|71=1=a=,.

tan a =

Clearly, the point representing z= 1+ iis P(A, 1), which Lies in the first quadrant.

T
arg(z)=0=a=7.
Hence, |z| = V2 and arg (z) = a

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14. 2
(4) Let z = —V3 + i. Then, |z| = (53 +12%= 44 = 2, 3
=
Let a be the acute angle given by 2
o | Im (z) =] 1 1 1 E
an a= = = a=-. f
Re (2) 13 v3 6 8
Now, z = (—V3 + i) is represented by the point P(—V3, 1), which Lies in the second El
quadrant. a
(m=0=(r-0)=(n-5=% 3
arg (z) = @ = (n-a) =(nx- =) =—.

cn 6) 6 =
Hence, |z| = 2 and arg (z) = m 3
2
=
n
jo)
=
5
®
®
3
£
=
oO
un

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15. > 2

(ús) Let z = -1 — iV3. Then, |z| = | (-1)? + (-V3) = V4 =2. $
Let a be the acute angle given by Ei
ld
m (z) -v3 TU ®
tana= Ip == 1=% a=3 a
Now, Z = (-1 = iv3) is represented by the point P(-1, -v3) , which lies in the 2
third quadrant. a
TE — 27 =
_ _ u _ Fe _ oo
arg (z)=@=(a n=(5 nm) => 5
Hence, |z| = 2 and arg (z) = = 3
2
5
jo)
=
5
®
®
3
£
o
un

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16. Convert the complex number (1 + iV3) into polar form:
The given complex number is z = (1 + iv3).
Let its polar form be z = r( cos 0 + isin 8).

Now, r = [xl = [12 + (V3) = V4 = 2.

Let a be the acute angle, given by
Im (z) V3 T
= = V3 =—,
Reel l1l LE
Clearly, the point representing z = (1+iV3)is P(1, V3) ‚which lies in the first
quadrant.

tana = |

arg (z) =@=a=-—.

ala

Thus, r = |z| = 2 and 8 = arg (2) = 5.

Hence, the required polar form of z = (1 + iV3) is 2 ( cos at isin 5) .

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17. Convert the complex number (2N3 — 2i) into polar form.
The given complex number is z = (2V3 — 2i).
Let its polar form be z = r( cos 6 + isin 8).

Now, r = |z| = „|(2V3)” + (22? = VI2 ¥4 = VI6 = 4.

Let a be the acute angle, given by

Im (z) | —2 | ER T
= a=-.

Re(z)| I2v3l V3 6

Clearly, the point representing z = (2V3 — 2i) is P(2V3,-2) „which lies in the

fourth quadrant.

tana =

—T
arg (Z)=0=-a=
Thus, r = |z| = 4 and O = arg (z) =<.

Hence, the required polar form of z = (2V3 — 21) is

4| cos (=) + ¿sin E = 4(cos = isin 2).

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18. Convert the complex number (=2 + 2iv3) into polar form.
The given complex number is z = (—2 + 2iv3) . Let its polar form be z =
r(cos 8 + i sin 8).

Now, r= [al = C2 + (2V8)" = VAT TZ = VIS = 4.

Let a be the acute angle, given by
Im (z 2v3 T
@|_ = 5 Seo
Re (z) —2 3
Clearly, the point representing z= (-2 + 2iV3) is P(-2,2V3) ; wnicn lies in the
second quadrant.

tana =

ar; @)=0=(n-a)=(n-2) ==

8 a ma = i gyi ge

Thus, r = |z| = 4 and 0 = =,
Hence, the required polar formás z = 4 ( cos = +isin =) €

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19. (1470,
Convert ona Into polar form.
_ (470 _ (+7) _ (470 | G+4ù
ipa (2-1? : (4+12-4i) o (3-40 © (+40
¿M4 (1+7i)(3+4i) z ( 5+25i la E + i) .

(9+16) 25

Let its polar form bez = r( cos 8 + isin 8). Now, r = |z| = Y(-1)? + 12 = v2.
7

Im (2) E lala
Re (o) =1=a=:

Clearly, the point representing z = (-1+4 à) is P(—1,1) , which Lies in the second
quadrant.

Let @ be the acute angle, given by tana =

need = 8=(T-a)= (r-7
Thus, 1 = [71 = V2 and =
Hence, the required polar form is z = V2 ( cos = +isin =) .

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20. Convert the complex number on fen into polar form.

1 @)_ 4) _G-D_f1 1,
am an an zie 21).

Let tts ote JE bez = r(cos@+isin@).

Now, r = |z| = 4425 a4
2 u arg vz

Im (2)
Re (z)

Letz=

1 2.
af]

Clearly, the point representing z = G 2 i) is P E 3) ‚which Lies in the fourth

Let a be the acute angle, given by tana = i

T
ri

2 2
auadrant. arg(z)=0=-a=
Thus, 1 = || = 5 and 6 = E

Hence, the required polar form is

z= {cos (7) + sin (<<) a, (eos; - isin À)
cos isin (= ))}, Ue, (cos ¿—isin 7).

ua
E

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21. Convert the complex ne into polar form.
(cos Fri sin 3)

_ G-1) _ G-1) _ 2-0
ar (cos 2+i sin) E an (1+/3)

2(i-1) „a= al (9,
"ara (1 —iv3) 2 u 2 a $

Let its polar form be z = r(cos O + isin 6).
Ea 2 = 2
Now, 1? = |7[2 = [en „use } =!-2=r=22.

4 4 4

Let a be the acute angle, given by

Im(),_,(/3+1) 2 6849. (+)

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tana =| =] = =
Re (2) 2 *(V3-1) (3-1) ces
v3.
u (tan £+tan?) al: wm) Sn Sn
> tana = a tan Fan?) = tan +5) = MR, PS
Clearly, the point representing the given complex number is P (E ar Bet) ‚ which
lies in the first quadrant.
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22. (2) =0 TE 51

= Ss = —>0=
PE a 12 12
Thus, r = |z| = V2 and 0 = 7.

Y har _ Sms
Hence, the required polar form is z = V2 ( cos > + isin =) .

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23. Express the complex number (1. — i) in polar form.
The given complex number is Z = —1—i.

Let its polar form be z = r( cos 8 + i sin 0).

Now, r = |z| = ED? + (ED? = v2.
Let @ be the acute angle, given by

Im (z)
Re (z)
Clearly, the point representing the complex number Z = —1— its P(-1,-1), which
lies in the third quadrant.

arg (2) = @ = ~(n- a) =~ (n-T) =

t S| 1 E
ana — = = a==
= 4

Thus, 7 = |g) = V2 and 0 =

Hence, the required polar form of z = (—1 — i) is given by
= V2{ cos (=) +isin (=) Le, V2 (cos o isin =) A

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24. Express the complex number (V3 — à) in polar form.
The given complex number is z = (—V3 — i).
Let its polar form be z = r( cos 0 + isin 8).

Now, r = |z| = (43) + (192 = v4 = 2.

; _ [Im (2)
Let a be the acute angle, given by tana = | RES logis mE

Clearly, the point representing the complex number z = (- V3 — a is CE -1) 3
which Lies in the third quadrant.

TC —57

arg(z)=09=-(n-a)= -(x-=) = ——

TT

T

thus, r = |z| = 2 and 0 =
Hence, the polar form of Z = a v3 — i) is given by

z= 2{cos (= )+isin (= =), Gex 2 (cos = — isin =).

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25. 142i,

Convert the complex number into polar form.

(142i) (+30) _ (1420430 _ —5+5i _ (1,1,
ae ar 10 7 ( 2+* i)
Let its polar form bez = r(cos 0 + isin 0).

cr YO E

Let a be the acute angle, given by tana=|

Let z =

Im (2)
Re (z)

pe EVEN
-(1/2)
. , -1,1.\, -11 et, JEAN
Clearly, the point representing z = E +3 li) is P E 2 3 , which Lies in the second
quadrant.

a
¡ISt5uez

TC Im
arg (z) = 0=(m-a=(n-7)=%
Thus, F = 171 ana =
1

; ha ose 3x, cn 37
Hence, the required polar form is z = Gl cos +isin =) 5

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26. Express the complex number sin = +i (1 — cos 5) in polar form.
a x
Let z = sin E+i(1- cos 2).
Let its polar form be z = r(cos 0 + isin 6) :
2

Now, r? = |z| = sin? 2+(1- cos 2) = (sin? =+cos 22) +1- 2 cos =

Sent =2 (1 cos 5) = Asin? = amsg2osimi,

5 10 10

Let a be the acute angle, given by
Im (z) 1- cos = 2sin?= =

land ”=|= = tan a=—
Re (2) sin= 2sin = cos E 10 10

tana = |

Clearly, the point representing Z lies in the first quadrant as x > 0 and y > 0.

=) are, =2sin = = Y
arg (z) =0=a=—. Thus, r 2 sin „and 10:

4 4 A nm ua i ov Tr
Hence, the required polar form is 2 sin 10 ( cos 5 + isin =) :

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27. Convert 4(-cos 300° + i sin 300°) into Cartesian form:
4( cos 300° + i sin 300°) = 4[ cos (360° — 60°) + i sin (360° — 60°)]
= 4[ cos (-60°) + i sin (-60°)]

= 4[ cos 60° = i sin 60°] =4 (1-54) = (2= 2443).

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28. For any complex numbers 2,24 and 22 prove that:
() arg (z) = — arg (z)

(@ arg (2122) = arg (21) + arg (22)

(Gi) arg (2122) = arg (z1) — arg (22)

(6) arg (2) = arg (21) - arg (22)

() Leez = r(cos 6 + isin 0). Then, |z| = r and arg (z) = 6.
Now, Z=r(cos@+isin@) > z=rcos 0 + i(r sin 0)
> Z=rcos 0 —i(r sin 0) = r( cos 6 — isin 0) = r{ cos (—@) + i sin (-0))

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> |z| = 7 and arg (2) = -0 = —arg(z).
Hence, arg (Z) = — arg (z).
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29. (it) Let zy = 174 ( cos 0, +i sin 01) and z, =r2(cos 0, + i sin 0).
Then, |z,| = T1, arg (Z1) = 0 and 1,,| = 12, arg (22) = 02.
Z122 = r,(cos 6, + isin 6,) -r,( cos 8, + isin 02)
=rırz{(cos 6, cos 6, — sin 6, sin 02) + ¿(sin 0, cos 6, + cos 6, sin 0,)}
= rırz{ cos (0, + 02) + i sin (0, + 02))
> arg (2122) = (0, + 02) = arg (21) + arg (zz).

(tit) Let zı =r,(cos 0, + isin 04) and z, = r¿( cos 0), + ¿sin ,).
Then,

Z2 = T2 COS 02 + ı(r, sin 82) = r2 cos 0) — i(r, sin 82)

> Z2 = r2[ cos (-0,) + isin (-0,)}.
Z4Z2 = r,(cos 6, + isin 6) - r2{ cos (—@2) + isin (—@2)}
= r1r2( cos 6, + isin 6,)( cos (-0,) + i sin (-0,)}
= r¡r2[ cos {0} + (-0,)} + i sin {6, + (—O2)}]
= 1 112{ cos (01 — 62) + isin (0, — 62)).

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30. (tv) Let zy =r,(cos 0; +isin 01) and z2 = r2(cos 67 + isin 0).
Then, |z,| = 71, |z,| = T2, arg (21) = 01 and arg (Zz) = 02.
Zı 14(cos@,+isin@,) (cos @2—isin 62)
Z2 T2(cos 02 +isin 0) (cos @,—isin 62)
ay { cos @- cos 0 — sin 6 - sin 0) + i( sin 0, : cos 62 — cos 0, sin 2)

(cos28;, + sin?62)

= (cos (0, — 62) + à sin (0, — 62)}
2

T2

> arg (#) = (0, - 62) = arg (21) - arg (22).
Hence, arg E) = arg (z 1) — arg (22).

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31. tf Zy and Zz are conjugate to each other, find the principal argument of (=21Z2) .
AS Z1 and Z2 are conjugate to each other L.e., Z2 = Zy, therefore, Z1Z2 = 2424 = Iz1l?
arg (—2122) = arg (—|z,|*) = arg [negative real numberl= nr

Let z be any non-zero complex number, then find the value of arg (z) + arg (2).
arg (z) + arg (Z) = arg (zZ)= arg (|z|*) = arg [positive real numberl= 0

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