Complex Numbers 3A
1 a | z  3| 3| z 5 |
| x  iy  3 |  3 | x  iy  5 |
| ( x  3)  iy |  3 | ( x  5)  iy |
| ( x  3)  iy |2  32 | ( x  5)  iy |2
 ( x  3) 2  y 2  9[( x  5)2  y 2 ]
 x 2  6 x  9  y 2  9[( x 2  10 x  25  y 2 )]
 x 2  6 x  9  y 2  9 x 2  90 x  225  9 y 2
 0  8 x 2  96 x  8 y 2  216 (8)
 x 2  12 x  y 2  27  0
 ( x  6)2  36  y 2  27  0
 ( x  6)2  y 2  9  0
 ( x  6)2  y 2  9
The Cartesian equation of the locus of z is ( x  6)2  y 2  9.
This is a circle centre (6, 0), radius = 3
b
| z  3 |  4 | z  1|
| x  iy  3 |  4 | x  iy  1|
| x  3  iy |2  16 | x  1  iy |2
( x  3)2  y 2  16(( x  1)2  y 2 )
x 2  6 x  9  y 2  16( x 2  2 x  1  y 2 )
 16 x 2  32 x  16  16 y 2
15 x 2  38 x  15 y 2  7  0
x2 
38
7
x  y2   0
15
15
2
19  19 2
7

2
x


  2 y  0
15  15
15


19 
256

2
x   y 
15 
225

16
 19 
Circle centre   , 0  radius
15
 15 
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1
1 c
| z  i | 2 | z  i |
| x  iy  i |  2 | x  iy  i |
| x  i( y  1) |2  4 | x  i( y  1) |2
x 2  ( y  1) 2  4[ x 2  ( y  1) 2 ]
x 2  y 2  2 y  1  4( x 2  y 2  2 y  1)
 4 x2  4 y 2  8 y  4
3 x 2  3 y 2  10 y  3  0
x2  y2 
10
y 1  0
3
2
5  25

x   y    1  0
3
9

2

5  16

x  y  
3
9

5
4

Circle centre  0,   radius
3
3

2
d
| z  2  7i |  2 | z  10  2i |
| x  iy  2  7i |  2 | x  iy  10  2i |
| ( x  2)  i( y  7) |2  4 | ( x  10)  i( y  2) |2
( x  2) 2  ( y  7)2  4[( x  10) 2  ( y  2)2 ]
x 2  4 x  4  y 2  14 y  49  4[ x 2  20 x  100  y 2  4 y  4
3 x 2  84 x  3 y 2  30 y  363  0
x 2  28 x  y 2  10 y  121  0
( x  14) 2  142  ( y  5)2  52  121  0
( x  14) 2  ( y  5)2  100
Circle centre 14,  5  radius10
e
| z  4  2i |  2 | z  2  5i |
| x  iy  4  2i |  2 | x  iy  2  5i |
| ( x  4)  i( y  2) |2  4 | ( x  2)  i( y  5) |2
( x  4) 2  ( y  2) 2  4[( x  2)2  ( y  5) 2 ]
x 2  8 x  16  y 2  4 y  4  4[ x 2  4 x  4
 y 2  10 y  25]
3 x 2  24 x  3 y 2  36 y  96  0
x 2  8 x  y 2  12 y  32  0
( x  4) 2  16  ( y  6)2  36  32  0
( x  4) 2  ( y  6)2  20
Circle centre  4, 6  radius 20  2 5
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1 f
| z | 2 | 2  z |
 2 | 1|| z  2 |
| x  iy |  2  1 | x  iy  2 |
x 2  y 2  4(( x  2)2  y 2 )
x 2  y 2  4( x 2  4 x  4  y 2 )
3x 2  16 x  3 y 2  16  0
16
16
x2  x  y 2   0
3
3
2
8  64

2 16
x    y  0
3
9
3

2
8
16

2
x   y 
3
9

4
8 
Circle centre  , 0  radius
3
3 
2 a
 z  π
arg 

 z 3 4
arg z  arg( z  3) 
π
4
arg z  arg( z  (3)) 
π
4
arg z  
arg( z  (3))  
π
4
π
  
4
P lies on an arc of a circle cut off at
A(3, 0) and O(0, 0)
Angle at the centre is twice the angle at the
  
circumference 
π
2
 3 3
It follows that the centre is at   , 
 2 2
3
and the radius is
2
2
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2 b
 z  3i  π
arg 

 z4  6
arg( z  3i)  arg( z  (4)) 
π
6
arg( z  3i)   .
arg( z  (4))  
π
  
6
Arc of a circle from (4,0) to (0,3)

 The centre is at



 43 3 3 4 3 
,
 
 , though you do not need to calculate this for a sketch. 
2
2 


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2 c
 z  π
arg 

 z 2 3
arg z  
arg( z  2)  
π
  
3
As our diagram has    , we have P on the wrong side of
the line joining O or  .
We want the arc below the x -axis.
Redrawing:
arg z  
arg( z  2)  
Hence arg z  arg( z  2) 
becomes    ( ) 
  
π
3
π
3
π
3
Arc of a circle, ends 0 and 2, subtending angle

 The centre is at

π
3
1 
2 3

 1, 
 radius 3 not needed
3


to be calculated for a sketch 

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 z  3i  π
arg 

 z 5  4
2 d
arg( z  3i)  arg( z  5) 
π
4
arg( z  3i)  
arg( z  5)  
π
  
4
But    , we have P on the wrong side of the line
joining 3i and 5.
arg( z  3i)  
arg( z  5)  
π
  ( ) 
4
π
  
4
(Arc of circle centre (1, –1) radius
17 not needed for sketch)
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arg z  arg( z  2  3i) 
2 e
π
3
arg z  arg( z  (2  3i)) 
π
3
arg z  
arg( z  (2  3i))  
π
  ( ) 
3
π
  
3
 2 3 9 2 3 
Arc of circle, centre at 
,
,
6 
 2
this need not be calculated for your sketch.
 z  4i  π
arg 

 z4  2
f
arg( z  4i)  arg( z  4) 
π
2
arg( z  4i)  
arg( z  4)    arg( z  (4i))
π
  
2
π
  
2
The locus is an arc of a circle, ends at –4 and 4i, angle subtended being
π
2
 It is a semi-circle.
(Circle arc has centre (–2, 2), radius 2 2 )
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3 a | z  1  i |  2 | z  4  2i |
| x  iy  1  i |  2 | x  iy  4  2i |
| ( x  1)  i(y  1) |  2 | ( x  4)  i( y  2) |
| ( x  1)  i(y  1) |2  22 | ( x  4)  i(y  2) |2
 ( x  1)2  ( y  1) 2  4[( x  4)2  ( y  2) 2 ]
 x 2  2 x  1  y 2  2 y  1  4[( x 2  8 x  16  y 2  4 y  4]
 x 2  2 x  1  y 2  2 y  1  4 x 2  32 x  64  4 y 2  16 y  16
 0  3 x 2  30 x  3 y 2  18 y  64  16  1  1
 3x 2  30 x  3 y 2  18 y  78  0
 x 2  10 x  y 2  6 y  26  0
 ( x  5)2  25  ( y  3)2  9  26  0
 ( x  5)2  ( y  3)2  25  9  26
 ( x  5)2  ( y  3)2  8
Therefore the locus of P is a circle centre (–5, 3). (as required)
b radius  8  4 2  2 2
The exact radius is 2 2.
4 a arg( z )  arg( z  4) 
   
π
4
π
, where arg( z )   and arg( z  4)  
4
from ΔAOP,
ˆ   
APO
ˆ   
 APO
ˆ  π
 APO
4
The locus of points P is an arc of a circle cut off at (–4, 0) and (0, 0), as shown below.
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4 b
Therefore the centre of the circle has coordinates (–2, 2).
c r  22  22  8  4 2  2 2
Therefore, the radius of C is 2 2.
d The Cartesian equation of C is ( x  2) 2  ( y  2) 2  8.
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4 e Finite area  Area of major sector ACO  Area ΔACO
1
π 1

( 8)2  2π    (4)(2)
2
2 2

1 
π
 (8)  2π    4
2 
2

 3π 
 4   4
 2 
 6π  4
Finite area bounded by the locus of P and the x-axis is 6π  4.
b, c, d
Method (2):
 z 
arg z  arg( z  4)  arg 

 z4
 x  iy 
 arg 

 x  iy  4 
 x  iy 
 arg 

 ( x  4)  iy 
 x  iy
( x  4)  iy 
 arg 


 ( x  4)  iy ( x  4)  iy 
 x ( x  4)  iyx  iy ( x  4)  y 2 
 arg 

( x  4)2  y 2


 x( x  4)  y 2   y ( x  4)  yx  
 arg 
 i
2
2 
2
2 
 ( x  4)  y   ( x  4)  y  
 x 2  4 x  y 2   xy  4 y  xy  
 arg 
i
2
2 
2
2 
 ( x  4)  y   ( x  4)  y  
 x 2  4 x  y 2  

4y
 arg 

i


2
2
2
2 
 ( x  4)  y   ( x  4)  y  
 z  π
Applying arg 
 
 z4 4



4y
( x  4)2  y 2
x 4 x  y
( x  4)2  y 2
2
2
  tan  π   1
  4 
4y
1
x  4x  y2
2
 4 y  x2  4 x  y 2
 0  x2  4x  y2  4 y
 ( x  2)2  4  ( y  2)2  4  0
 ( x  2)2  ( y  2)2  8
 ( x  2)2  ( y  2)2  (2 2)2
C is a circle with centre (–2, 2), radius 2 2 and has Cartesian equation ( x  2)2  ( y  2)2  8.
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5 a Curve F is described by z  2 z  4 . First, note that z can be written as z = x + iy:
x  yi  2x  2 yi  8 . Next, group the real and imaginary parts
x  yi  2  x  4   yi . Square both sides
x  yi  22  x  4   yi
2
2
x 2  y 2  4  x  4  4 y 2
2
x 2  y 2  4  x 2  8 x  16   4 y 2
x 2  y 2  4 x 2  32 x  64  4 y 2
4x 2  32x  64  x 2  4 y 2  y 2  0
3x 2  32x  3y 2  64  0
32
64
x2  x  y 2 
0
3
3
Completing the square for x
2
64  8 
 16 
2
 
x   y 
3
9  3

2
8
 16 
Thus we see that F is a circle centred at   , 0  with radius r 
3
 3 
b
8
8
 16 
c The circle is centred at   , 0  and its radius is r  . This means that it stretches out from 
3
3
 3 
8
8
8
to along the imaginary axis. Thus 
Im  z 
3
3
3
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11
6 We are given curve defined by z  8  2 z  2  6i . To visualise this, express z as real and imaginary
parts and square both sides
x  8  yi  2 x  2  yi  6i
2
x  8  yi  22 x  2  yi  6i
 x  8
2
2
 y2  4  x  2  4  y  6
2
2
x 2  16 x  64  y 2  4 x 2  16 x  16  4 y 2  48 y  144
3 x 2  3 y 2  48 y  96  0
x 2  y 2  16 y  32  0
x 2   y  8   64  32  0
2

x 2   y  8   32  4 2
2

2
 
So this curve is a circle centred at 0,8 with radius r  4 2 . Now the largest and smallest values of
arg  z  will be found at the points of tangency of the circle to the lines going through the origin.
These are shown below as z1 and z2.
We can calculate the distance x from the origin to A using Pythagoras Theorem:
x 2  r 2  82
x 2  64  32
x4 2r
So the triangle created by the origin, z1 and the centre of the circle is a right-angled isosceles triangle,
π
π
π
so the angle ⊲ COA  . Similarly, ⊲ COB  . Thus we conclude that arg  z1   and
4
4
4
3π
π
3π
arg  z2  
. So for any z lying on this circle we have
arg  z 
4
4
4
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12
 w  8i  π
7 a We want to sketch the curve S satisfying arg 
  . We have
 w 6  2
π
 w  8i 
arg 
  arg  w  8i   arg  w  6       , where arg w  8i   and arg  w  6    .
2
 w 6 
π
Since the constant angle is , S is a semicircle from 0,8 anticlockwise to 6,0 but not
2
including these two points.


 




 
b The centre of this semicircle lies in the middle of the line connecting 6,0 and 0,8 , i.e. at
 3,4 . The radius can be found be using Pythagoras Theorem:
r 2  32  42  25
r 5

 
2

2
Thus the Cartesian equation for S can be written as x  3  y  4  25, x < 0, y  0 .


 
Remember to specify the range of x and y. Here the inequalities are strict since 6,0 and 0,8
are not included in the curve.
c The argument of an imaginary number z is the angle between the line connecting z to the origin
and the real axis.
For curve S the smallest such angle is for z  8i and the largest for z  6 . Remember that the
π
endpoints are not included in the curve, so we have < arg  z  < π
2
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13

7 d The point furthest to the left is 8  4i , so the smallest possible value of Re z is -8. The
endpoints of the semicircle are not included in the curve, so we need to use a strict inequality for
the largest value of Re z . Thus 8 Re  z  < 0 .



3π
, z  3 . Let L1 be the half-line satisfying arg z  1   and L2
4
3π
be the half-line satisfying arg z  3   . From the initial equation we have    
4
8 We have arg  z  1  arg  z  3 


Now considering the triangle APB we see that
ˆ
ˆ  APB
ˆ  DAP
PBA
ˆ  PBA
ˆ  DAP
ˆ      3π
APB
4
3π
4
So the locus will be an arc going anticlockwise from A to B:
So, as  and  vary, the angle APB remains constant at
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8 (continued)
Now we know that the centre of this circle lies on the perpendicular bisector of the line segment
connecting A and B, which has equation x  1. Let C be the centre of this circle.
ˆ  3π , so BCA
ˆ  2π  2 ADB
ˆ π.
We know that ADB
4
2
So ACB is an isosceles, right-angled triangle.
So we have:
r 2  r 2  42
r2  8
r2 2
Now, using Pythagoras Theorem again, on triangle BEC we have that
CE 2  22  r 2
CE 2  4
CE  2
So the centre has coordinates C  (1,2) and the Cartesian equation of this locus can be written as
 x 1   y  2
2
2
 8, y  0 .
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9 a
By considering the triangle APB, we have that
ˆ  APB
ˆ  OAP
ˆ
PBA
ˆ  PBA
ˆ  OAP
ˆ
APB
ˆ  PBA
ˆ π
OAP
4
Moreover, we know that angles in the same segment of a circle are equal, so we’re looking for all
π
numbers z for which arg  z  2   arg  z  5  
4
 z 2 π
Thus the equation describing this locus is arg 

 z 5 4
b
Similar to example a, we have
π
arg  z  i   arg  z  4i  
6
 z i  π
So arg 

 z  4i  6
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9 c
Using the same the same techniques as for part a and b we have that the locus can be described as
2π
arg  z   6  i    arg  z  1  2i   
3
2π
arg  z  6  i   arg  z  1  2i  
3
 z  6  i  2π
arg 

 z  1  2i  3
10 a We have z  3  3 z  5 . By representing z as real and imaginary parts and squaring both sides of
the equation we see that:
x  3 yi  3 x  5  yi
2
x  3 yi  9 x  5  yi
 x  3
2
2
 y 2  9  x  5  9 y 2
2
x 2  6x  9  y 2  9x 2  90x  225  9 y 2
8x 2  96x  216  8y 2  0
x 2  y 2  12x  27  0
as required.
b The above equation can be rewritten as follows:
2
 x  6   36  y 2  27  0
 x  6  y 2  9
2
 x  6   y 2  32
2
 
So the equation describes a circle centred at 6,0 with radius r  3
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

π
π
and that z1 ÎC . If we write z1  r cos  isin  where   , we see
6
6
that z1  23 r  12 ri . Moreover, we know that z1 lies on the circle, so if we write z1  x  yi , x and y
10 c We have that arg  z1  


2
must satisfy x  6  y 2  32 . Comparing the two expressions for z1 , we obtain
3
1
r , y  r . Substituting these values into the circle equation we have:
2
2
2
  1 2
3
r  6    r   9
2
 2 
x



3 2
1
r  6r 3  36  r 2  9
4
4
2
r  6r 3  27  0
r  3 3 
2
0
r 3 3
π
π

Thus we can write z1  3 3  cos  i sin 
6
6

11 a We have the locus of points P satisfying z  z1  k z  z2 . Moreover, we know that AP  2BP ,
A   0,6  , B   3,0  . Thus we can write z  6i  2 z  3 .
b Write z  x  yi and square both sides of equation derived in part a:
x  yi  6i  2 x  3  yi
2
x  yi  6i  4 x  3  yi
2
x 2   y  6   4  x  3  4 y 2
2
2
x 2  y 2  12 y  36  4 x 2  24 x  36  4 y 2
3 x 2  24 x  3 y 2  12 y  0
x2  y2  8x  4 y  0
as required.

 
2

2
 
2
c The equation for circle C derived in part b can be written as x  4  y  2  20  2 5 . This


means the circle is centred at 4,2 and has radius r  2 5 . We are given the locus of points w


satisfying arg w  6   and  passes through the centre of the circle. The centre is at point
c  4  2i and we know that, since the centre lies in the 4th quadrant,
Thus we can write tan  2π     

Im  c 
 tan  2π    .
Re  c 

1
and so, since  Î 0,2p , we have that
2
 1
2π    tan 1     0.46
 2
  5.82
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

11 d We know that Q satisfies both arg w  6   and m  n  b , since it lies on the intersection of the
y
1
line and the circle. Thus, writing q  x1  y1i we have 1    x1  2 y1 .
x1
2
Substituting this into the circle equation, we obtain:
4 y12  y12  16 y1  4 y1  0
5y12  20 y1  0
y1  y1  4   0
y1  0 or y1  4


y1  0 leads to x1  0 , so the origin. Thus we take y1  4 and x1  8 . So Q  8,4 .
Challenge
The equation z  a  z  a  b describes all points P for which the sum of distances from a and –a is
equal to b. According to the graph above, we have m  n  b . This is exactly the definition of an
ellipse with foci at a and –a and the major axis of length b.
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