Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

www
...
com

COMPLEX NUMBERS
1
...


2
...


3
...


4
...

(b) In the set of complex numbers, there is no meaning to the phrases one complex is
greater than or less than another i
...
If two complex numbers are not equal, we say they
are unequal
...


5
...
They are of the
form a+ib, a-ib
...


7
...

9
...


cisθ1 cisθ2 = cis(θ1 + θ2 ) ,
a 1 + ib 1
a 2 + ib 2

=

(a a
1

2

cis θ 1
cis θ2

) (

= cis(θ1 – θ2 ),

+ b1b2 + i a2b1 − a1b2
a2 + b2
2

1
= cosθ − isinθ
...

11
...
Rule for choosing the principal amplitude
...
sakshieducation
...
sakshieducation
...


If α is the principle amplitude of a complex number, general amplitude = 2nπ + α where
n∈Z
...
i) Amp (Z1 Z2 ) = Amp Z1 + AmpZ2
 Z1 
 = AmpZ 1 − AmpZ 2
 Z2 

ii) Amp

iii) Amp z + Amp z = 2π (when z is a negative real number) = 0 (otherwise)
14
...


15
...

2

16
...

4

www
...
com

www
...
com

Very Short Answer Questions

1
...


If z1 = ( 3, 5 ) and z2 = ( 2, 6 ) find z, z2

Solution: Given z1 = 3 + 5i

z2 = 2 + 6i

z1 − z2 = ( 3 + 5i )( 2 + 6i ) = 6 + 280 C + 30i 2 = 6 − 30 + 28i
z1
...


Write the additive inverse of the following
(i) Additive inverse of

Additive inverse is

(−

(

3, 5

3, − 5

)

)

(ii) Additive Inverse of ( −6, 5) + (10, − 4 )
Given complex no is (4,1)
Additive inverse is

( −4, − 1)

www
...
com

www
...
com
(iii) Additive inverse of ( 2, 1) ( −4, 6 )
Let z1 = 2 + i z2 = − 4 + 6i
z1 z2 = ( 2 + l ) ( −4 + 6i ) = − 8 +8i − 6i 2
= − 2 + 8i

Additive inverse is (-2, -8)

4
...


12  12 
i =  1, 
5
 5 

If z = cos θ + i sin θ then find z −

1
z

Solution: z = cos θ + i sin θ
1
1
cos θ − i sin θ
=
×
z cos θ + i sin θ cos θ − i sin θ
= cos θ − i sin θ

⇒

z−

1
= 2i sin θ
z

www
...
com

www
...
com
6
...
sakshieducation
...
sakshieducation
...
Write the following complex numbers in the form A + iB
...
sakshieducation
...
sakshieducation
...
sakshieducation
...
sakshieducation
...
Write the conjugate of the following complex numbers
...
Simplify
i) i 2 + i 4 + i 6 +
...
i 7 + i 2 1 + i 4 ⋅ ( −i )
Sol:

26

i) i 2 + i 4 + i 6 +
...
2 + i 2
...
i 2
...
2n terms + i 2
= 0 − 1 = −1

(

)

ii) i18 − 3
...
sakshieducation
...
sakshieducation
...
Find a square root for the following complex numbers
...
(1)
| a + ib | = | 7 + 24i |

Squaring on both sides,
| a + ib |2 =| 7 + 24i |
a 2 + b2 = 49 + 576
a 2 + b2 = 625 = 25
...

ii) −47 + i8 3
Let the square root of z be a + ib,
(a + ib) 2 = −47 + i8 3
a 2 − b 2 = −47, 2ab = 8 3
| a + ib |=| −47 + i8 3 |

www
...
com

www
...
com
Squaring on both sides,
a 2 + b 2 = (−47) 2 + (8 3)2
= 2209 + 192 = 2401
a 2 + b 2 = 49
a 2 − b 2 = −47
2a 2 = 2
a2 = 1
a = ±1
2b 2 = 96
b 2 = 48
b = ±4 3
a + ib = ± (1 + 4 3i)

11
...

i) 5 + 3i
Sol:

ii) –i

iii) i −35

i) z = 5 + 3i

Let a + ib be multiplicative inverse then
(a + ib)z = 1
z=

1
1
or a + ib =
a + ib
z

a + ib =

z
(zz)

a + ib =

z
5 − 3i 1
=
= ( 5 − 3i)
2
5+9
14
|z|

ii) z = –i
Let a + ib be multiplicative inverse then
(a + ib)z = 1
a + ib =

1
z

www
...
com

www
...
com
=

1
−i

=

i
−i ⋅ i

a + ib = i

iii) z = i −35
Let a + ib be multiplicative inverse then
(a + ib)z = 1
a + ib =

1
1
= −35 = i35
z i

(a + ib) = i35 = (i 2 )17
...
Express the following complex numbers in modules – amplitude form
...
sakshieducation
...
sakshieducation
...
sakshieducation
...
sakshieducation
...
Simplify –2i(3 + i)(2 + 4i)(1 + i) and obtain the modulus of that complex number
...
i) If z ≠ 0 find Arg z + Arg z
...

 z2 

iii) If z1 = –1 and z2 = i then find Arg 
Sol:

i) let z = x + iy ⇒ z = x − iy
Arg z = tan −1

y
 −y 
and Argz = tan −1 

x
 x 

Arg z + Argz = tan −1
= tan −1

y
 −y 
+ tan −1 

x
 x 

y
y
− tan −1  
x
x

=0

www
...
com

www
...
com
ii) z1 = –1 and z2 = –i
Arg(z1z 2 ) = Arg z1 + Arg z 2
0
−1
+ Tan −1
−1
0
π π
= π− =
2 2

= Tan −1

iii)z1 = –1 and z2 = i
z 
Arg  1  = Arg z1 − Arg z 2
 z2 
0
1
− Tan −1
0
−1
π π
= π− =
2 2

Tan −1

15
...

ii) If 3 + i = r(cos θ + i sin θ) then find the value of θ in radian measure
...
cis β then find the value of x2 + y2
...

z1
2z1 − z 2

v) If ( 3 + i)100 = 299 (a + ib) then show that a2 + b2 = 4
...
sakshieducation
...
sakshieducation
...
sakshieducation
...
sakshieducation
...
i) If z = x + iy and |z| = 1, then find the locus of z
...


ii) If the amplitude of (z – 1) is π/2 then find the locus of z
...


17
...


Sol:

i)z1 = 7 + 7i, z2 = 7 – 7i

Let A (7, 7),B(7, –7) be the points which represents above complex nos
...

Therefore equation of the line passing through (7,0) and having slope 0 is y = 0
...
sakshieducation
...
sakshieducation
...

Sol
...
are A(–9, 6)
Slope of AB =

B(11, –4)

6+4
10
−1
=
=
−9 − 11 −20 2

Equation of line AB:
−1
(x + 9)
2
2y − 12 = − x − 9
y−6 =

x + 2y = 3

18
...
sakshieducation
...
sakshieducation
...


 z −1  π
=
 z +1  4

iv) Arg 
z1 =
=

z − 1 (x − 1) + iy
=
z + 1 (x + 1) + iy

[(x − 1) + iy][(x + 1) − iy]
[(x + 1) + iy][(x + 1) − iy]

(x − 1)(x + 1) + y 2 + iy(x + 1 − x + 1)
=
(x + 1) 2 + y 2
Argz1 = Tan −1

2y
π
=
2
x + y −1 4
2

2y
=1
x + y2 − 1
2

2y = x 2 + y 2 − 1
x 2 + y 2 − 2y − 1 = 0

www
...
com

www
...
com
19
...


Sol:

Let A(2, 2), B(–2, –2), C( −2 3, 2 3 ) be points represents given complex numbers in the

argand plane
...


20
...
Given equation is of the form
Where S ( 4, 0 )

 12 
S '  , 0  and 2a =
 5 

12
= 10
5

SP+S1P =2a
10

⇒ a=5

SS’= 2ae
⇒ 4−

12
8
4
= 2 X 5e ⇒ = 10e ⇒ e =
5
5
5

21
...
sakshieducation
...

a − ib

www
...
com
a 2 − b2
2ab
= 2
+i 2
2
a +b
a + b2

∴ Real part =

a 2 − b2
a 2 + b2

Imaginary part =

2ab

...


If 4x + i(3x – y) = 3 – 6i where x and y are real numbers, then find the values of x and y
...
Equating the real and imaginary parts in the above

equation, we get 4x = 3, 3x – y = –6
...


23
...

Sol:

z = 2 – 3i ⇒ z – 2 = –3i ⇒ (z – 2)2 = (–3i)2
⇒ z2 + 4 – 4z = –9
⇒ z2 – 4z + 13 = 0
...
sakshieducation
...
sakshieducation
...

i) If (a + ib) 2 = x + iy , find x2 + y2
...

2 + cosθ + isinθ

3
rationalizing the dr
...
H
...
=
2

 6 + 3 cos θ   −3 sin θ 
x +y =
 +

 5 + 4 cos θ   5 + 4 cos θ 
2

2

2

36 + 9 cos 2 θ + 36 cos θ + 9sin 2 θ
=
(5 + 4 cos θ)2

www
...
com

www
...
com
=

45 + 36 cos θ
(5 + 4 cos θ)2

=

9(5 + 4 cos θ)
(5 + 4 cos θ)2

x2 + y2 =

9
5 + 4 cos θ

R
...
S
...


iii)If x + iy =
x + iy =

1
then show that 4x2 – 1 = 0
...
sakshieducation
...
sakshieducation
...

z+3

2+i
z+3

=

2+i
(x + 3) + iy

=

(2 + i)(x + 3 − iy)
(x + 3) 2 + y 2

=

(x + 3)2 + y i(x + 3 − 2y)
+
(x + 3)2 + y 2 (x + 3) 2 + y 2

u=

2(x + 3) + y
x + 3 − 2y
,v =
2
2
(x + 3) + y
(x + 3) 2 + y 2

2
...

Sol:

i) z = 3 + 5i
(z – 3)2 = (5i)2
z 2 − 6z + 9 = 25i 2
z 2 − 6z + 9 = −25
z 2 − 6z + 34 = 0
z3 − 6z 2 + 34z = 0
(z3 − 10z 2 + 58z − 136) + 4z 2 − 24z + 136 = 0
(z3 − 10z 2 + 58z − 136) + 4(z 2 − 6z + 34) = 0
∴ z3 − 10z 2 + 58z − 136 = 0

ii) If z = 2 − i 7 then show that
3z3 – 4z2 + z + 88 = 0
...
sakshieducation
...
sakshieducation
...

25
(1 − 2i)2

2−i
(if z = a + ib, z = a − ib )
(1 − 2i) 2

z=

=

2−i
(1 − 4) − 4i

2−i
−3 − 4i
(2 − i)(−3 + 4i)
=
(−3 − 4i)(−3 + 4i)
=

−6 + 8i + 3i + 4
9 + 16
−2 + 11i
=
25

=

z=

−2 + 11i
[Conjugate to z is z ]
25

3
...

a b

i) (x − iy)1/ 3 = a − ib
x − iy = (a − ib)3
x − iy = a 3 + ib3 + 3a(−ib)(a − ib)

⇒ x = a 3 − 3ab 2 & − iy = ib3 − 3a 2 bi
⇒ x = a 3 − 3ab 2 & y = b3 − 3a 2 b

www
...
com

www
...
com
y = b3 − 3a 2 b

x
y
= a 2 − 3b 2 and = b 2 − 3a 2
a
b

(

x y
− = 4 a 2 − b2
a b

 a + ib 

)

2

 a − ib 

2

ii) Write 
 −
 in the form x + iy
...

(1 + i)x − 2i (2 − 3i)y + i
+
=i
3+i
3−i
[(1 + i)x − 2i](3 − i) [(2 − 3i)y + i](3 + i)
+
=i
(3 + i)(3 − i)
(3 − i)(3 + i)
x(3 − i + 3i + 1) + y(6 + 2i − 9i + 3) − 6i − 2 + 3i + 12
=i
9 +1
i(2x + 2 − 7y − 3) 4x + 9y − 3
+
=i
10
10
4x + 9y − 3 = 0
2x − 7y − 3 = 10

Solving we get, x = 3, y = –1
...
sakshieducation
...
sakshieducation
...
i)Find the least positive integer n, satisfying 
 = 1
...


3

3

1 + i  1 − i 
ii) If 
 −
 = x + iy find x and y
 1 −i  1 + i 

3

3

1 + i  1 − i 

 −
 = x + iy
 1 −i  1 + i 
3

3

 (1 + i)(1 + i)   (1 − i)(1 − i) 

 −
 = x + iy
 (1 − i)(1 + i)   (1 + i)(1 − i) 
3

 2i   −2i 
  −

2  2 

3

−i − 1 = x + iy
x=0
y = −2

www
...
com

www
...
com
iii)Find the real values of θ in order that
(a) Real Number

z=
z=

3 + 2i sin θ
is a
1 − 2i sin θ

(b) Purely Imaginary Number
...

3+i 3−i

( x − 1)( 3 − i ) ( y − 1)( 3 + i )
x −1 y −1
+
=i⇒
+
=i
3+i 3−i
( 3 + i )( 3 − i ) ( 3 − i )( 3 + i )

www
...
com

www
...
com
(x − 1)(3 − i) (y − 1)(3 + i)
+
=i
9 +1
9 +1
3(x − 1) + 3(y − 1) i(1 − x + y − 1)
+
=i
10
10
3x + 3y − 6 − x + y
+
=i
10
10
3x + 3y − 6 = 0
y − x = 10

Solving we get, x = –4, y = 6
...
Simplify the following complex numbers and find their modulus
...
i) If (1 − i)(2 − i)(3 − i)
...
(1 + n 2 ) = x 2 + y 2
...
(1 − ni) = x − iy
Taking modulus both sides
...
|1 − ni | = | x − iy |
2 ⋅ 5
...
⋅ (1 + n 2 ) = x 2 + y2

www
...
com

www
...
com
ii) If the real part of

z +1
is 1, then find the locus of z
...

|z – 3 + i| = 4
| (x − 3) + i(y + 1) | = 4
(x − 3)2 + (y + 1)2 = 16
x 2 + y 2 − 6x + 2y + 10 = 16
x 2 + y 2 − 6x + 2y − 6 = 0

iv)If | z + ai | = | z − ai | then find the locus of z
...
sakshieducation
...
sakshieducation
...
If z = x + iy and if the point P in the Argand plane represents z, then describe
geometrically the locus of P satisfying the equations
i) |2z – 3| = 7
Sol:

i)|2z – 3| = 7
| 2(x) − 3 + 2yi |= 7
(2x − 3) 2 + 4y 2 = 7
4x 2 − 12x + 9 + 4y 2 = 49
4x 2 + 4y 2 − 12x − 40 = 0
x 2 + y 2 − 3x − 10 = 0
3



7

Centre  , 0  , radius =
2
2 
ii) |z|2 = 4 Re (z + 2)
| z |2 = 4 Re(z + 2)

x2 + y2 = 4(x + 2)
x2 + y2 – 4x – 8 = 0
Circle centre (2, 0)
...


iii)|z + i|2 – |z – i|2 = 2
x 2 + (y + 1)2 − x 2 − (y − 1)2 = 2
4y = 2
2y = 1 ⇒ 2y − 1 = 0

Line parallel to x-axis
...

| (x + (y + 4)i) | + | (x + (y − 4)i) |= 10

www
...
com

www
...
com
x 2 + (y + 4)2 + x 2 + (y − 4)2 = 10

(

x 2 + (y + 4) 2 = 10 − x 2 (y − 4)2

)

2

x 2 + (y + 4) 2 =
100 + x 2 + (y − 4) 2 − 20 x 2 + (y − 4)2

Solving we get
25x2 + 9y2 = 225 is ellipse
...

5

8
...

Sol:

i)|z1 + z2| = |z1| + |z2|
Squaring both sides
| z1 + z 2 |2 = (| z1 | + | z 2 |)

2

(z1 + z 2 )(z1 + z2 )
= | z1 |2 + | z 2 |2 +2 | z1 || z 2 |
z1 z1 + z 2 z2 + z1 z2 + z 2 z1
= | z1 |2 + | z 2 |2 +2 | z1 || z 2 |
z1 z2 + z 2 z1 = 2 | z1 || z 2 |

(x1 + iy1 )(x 2 − iy 2 ) + (x 2 + iy 2 )(x1 − iy1 )
2
2
= 2 x1 + y1 x 2 + y 2
2
2

Squaring on both sides we get

www
...
com

www
...
com
2
2
2
2
(x1x 2 + y1y 2 )2 = (x1 + y1 )(x 2 + y 2 )

(x1y 2 − y1x 2 ) 2 = 0
y 2 y1
=
x 2 x1
∴ Arg z1 − Arg z 2 = 0

ii) If z = x + iy and the point P represents z in the Argand plane and

z−a
=1
z+a

Re(a) ≠ 0 then find the locus of P
...


z−a
=1
z+a

| z − a | =| z + a |

Squaring on both sides
(x – a)2 + y2 = (x + a)2 + y2
4xa = 0
x=0
Parallel to y-axis
...
If

z3 are collinear
...

z1 − z 2
 z1 − z 2 

θ=0
∴ z1, z2, z3 are collinear
...
sakshieducation
...
sakshieducation
...
Show that the four points in the Argand plane represented by the complex numbers 2 + i,
4 + 3i, 2 + 5i, 3i are the vertices of a square
...


11
...

2 2
2

 −3 1 

7 7

A(–2, 7), B  ,  , C(4, −3), D  , 
 2 2
2 2
2

3 
1

AB =  −2 +  +  7 − 
2 
2

=

1 169
170
+
=
4 4
2
2

3 
1

BC =  4 +  +  −3 − 
2 
2

=

2

121 49
170
+
=
4
4
2
2

7 
7

CD =  4 −  +  −3 − 
2 
2

=

2

2

1 169
170
+
=
4 4
2

www
...
com

www
...
com
2

7 
7

AD =  −2 −  +  7 − 
2 
2

=

2

121 49
170
+
=
4
4
2

Slope of AC =

7 + 3 10 −5
=
=
−2 − 4 −6 3

7 1
−
2 2 =3
Slope of BD =
7 3 5
+
3 2

AC ⊥ BD
∴ ABCD is rhombus
...
Show that the points in the Argand diagram represented by the complex numbers z1, z2, z3
are collinear if and only if there exists real numbers p, q, r not all zero, satisfying
pz1 + qz2 + rz3 = 0 and p + q + r = 0
Sol:

pz1 + qz2 + rz3 = 0

pz1 + qz2 = –rz3
 pz1 + qz 2 

 (p + q) = − rz3
 p+q 

Now p + q = –r
pz1 + qz 2
= z3
p+q

⇒ z3 divides z1 and z2 is q : p ratio
...


www
...
com

www
...
com
13
...
O is origin
...

z1 z2 + z1z 2 = 0

Sol:

 z  z 
z1 z2 + z1z 2
= 0 ⇒  1 +  1  = 0
 z2  z2  
z 2 z2


⇒ Re al of

Or

z1
=0
z2

z1
is purely imaginary
z2

 z1 
 is k
...

z2

z  π
⇒ Arg  1  =
...
Determine the locus of z, z ≠ 2i, such that 
=0
...

i
...
x2 – 4x + y2 – 2y = 0 if and only if
(x – 2)2 + (y – 1)2 = 5
...
sakshieducation
...
sakshieducation
...

Hence the locus of the given point representing the complex number is the circle with (2, 1) as
centre and

5 units as radius except the point (0, 2)
...
Write z = −7 + i 21 in the polar form
...
We find that θ =
is such a solution
...
If the amplitude of 
 = , find its locus
...
sakshieducation
...
sakshieducation
...

2

Hence a = 0 and b ≥ 0
∴ x(x – 2) + y(y – 6) = 0 or
x2 + y2 – 2x – 6y = 0

…(1)

and 3x +y – 6 ≥ 0

…(2)

The points satisfying (1) and (2) constitute the arc of the circle x2 + y2 – 2x – 6y = 0 intercepted
by the diameter 3x + y – 6 = 0 not containing the origin and excluding the points (0, 6) and
(2, 0)
...


17
...


Sol:

Assume the general form of the equation of a circle in Cartesian coordinates as
x 2 + y 2 + 2gx + 2fy + c = 0,(g,f ∈ R)
...
Then
z+z
z−z
i(z − z)
= x,
=y=−
2
2i
2
x 2 + y 2 = | z |2 = zz

Substituting these results in equation (1), we obtain
zz + g(z + z) + f (z − z)(−i) + c = 0

i
...
zz + (g − if )z + (g + if )z + c = 0
...


...
sakshieducation
...
sakshieducation
...
Show that the complex numbers z satisfying z 2 + z 2 = 2 constitute a hyperbola
...
Substituting z = x + iy in the given equation z 2 + z 2 = 2 , we obtain the Cartesian form of the
given equation
...
e
...
e
...

Since this equation denotes a hyperbola all the complex numbers satisfying z 2 + z 2 = 2 constitute
the hyperbola x2 – y2 = 1
...
Show that the points in the Argand diagram represented by the complex numbers 1+ 3i,
4 – 3i, 5 – 5i are collinear
...

Slope of PQ =

3+3 6
=
= −2
1 − 4 −3

Slope of QR =

−3 + 5
= –2
4−5

∴ P, Q, R are collinear
...
Find the equation of straight line joining the points represented by (–4+3i), (2 – 3i) in the
Argand plane
...
e
...


www
...
com

www
...
com
21
...
If the amplitude of z is π/4,
determine the locus of P
...

4

z
z
π
π
=
and y = z sin =
⋅
4
4
2
2

Hence x ≥ 0, y ≥ 0 and x = y
...

∴ The locus of P is the ray

{( x, y ) ∈ R

2

}

| x ≥ 0, y ≥ 0, x = y ⋅

22
...

Sol: We note that

z−i
is not defined if z = 1
...

2
z −1
( x − 1) + y 2

i
...
, x 2 + y 2 − x − y = 0 and ( x, y ) ≠ (1, 0 )
...


www
...
com

z−i
is a
z −1

www
...
com
23
...
z has argument θ , 0 < θ <



π
and satisfy the equation z − 3i = 3
...

z


Sol
...
sakshieducation

---