Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
MATH 1180: 
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
Solutions to Assignment #1
1. (a) Prove using the “conjugate method”: |z1 + z2 + z3|  |z1| + |z2| + |z3|.
 Hint: Consider |z1 + z2 + z3|2 , and that (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca .
 |z1 + z2 + z3|2

+ z ) z

z 
= (z1 + z2 + z3) z1  z 2  z3
= (z1 + z2
3
 z2
1
3
= z1 z1  z1 z 2  z1 z3  z 2 z1  z 2 z 2  z 2 z3  z3 z1  z3 z 2  z3 z3

 
 
 
= z1 z1  z 2 z 2  z3 z3  z1 z 2  z 2 z1  z 2 z3  z3 z 2  z3 z1  z1 z3

= z1  z 2  z3
2
2
 |z1 + z2 + z3|2 = z1  z 2  z 3
2
2
2
  z z
1
2
2

 
 

 z1 z 2  z 2 z3  z 2 z3  z3 z1  z3 z1
 2 Re z1 z 2


 2 Re z 2 z 3


 2 Re z 3 z1


Re(z)  |z| , and so 2Re(z)  2|z|, hence the above equation implies:
But recall generally, that:
|z1 + z2 + z3|2  z1  z 2  z 3
2
2
2
 2 z1 z 2
 2 z 2 z3
 2 z 3 z1
And to simplify the expressions on the R.H.S., recall that generally: w1 w2  w1 w2 and w  w .
So we now have:
|z1 + z2 + z3|2  z1  z 2  z 3
2
 2 z1 z 2
 2 z 2 z3
|z1 + z2 + z3|2  z1  z 2  z 3
2
 2 z1 z 2
 2 z 2 z3
2
2
2
2
 2 z 3 z1
 2 z 3 z1
Observe the R.H.S. is now the expansion of a trinomial perfect square of the form given in the hint.
 |z1 + z2 + z3|2 
z
1
 z 2  z3

2
And, taking the positive square root of both sides (because modulus is always positive) gives:
|z1 + z2 + z3|
 z1  z 2  z3
…..shown.
■
(b) By using the polar form of two general complex numbers z1 and z2 show that:
arg(z1 z2) = arg(z1) + arg(z2) and arg(z1/z2) = arg(z1) – arg(z2)
 Let z1 = r1(cos1 + i sin1) and
z2 = r2(cos2 + i sin2)
Hence: z1 z2 = r1(cos1 + i sin1) r2(cos2 + i sin2)
= r1 r2 (cos1 + i sin1) (cos2 + i sin2)
= r1 r2 [(cos1 cos2 – sin1 sin2) + i (cos1 sin2 + sin1 cos2)]
= r1 r2 [(cos(1 + 2) + i (sin1 cos2 + cos1 sin2)]

z1 z2 = r1 r2 [(cos(1 + 2) + i sin(1 + 2)]
Observe the equation above shows a complex number z = z1 z2 in the form r(cos + i sin), where its
modulus r is given by r1 r2 and its argument (arg z) is given by  = 1 + 2 = arg(z1) + arg(z1).
Therefore, arg(z1 z2) = arg(z1) + arg(z2).
…..shown.
■
Now, for the quotient z1/z2:
z1
r cos 1  i sin 1 
 1
z 2 r2 cos  2  i sin  2 
Rationalizing:
z1
r cos 1  i sin 1  cos  2  i sin  2 
 1

z 2 r2 cos  2  i sin  2  cos  2  i sin  2 
1
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
z1 r1  cos 1 cos  2  sin 1 sin  2   i cos 1 sin  2  sin 1 cos  2 

z2
r2 cos 2  2  sin 2  2


z1 r1  cos(1   2 )  i sin( 1   2 )

z2
r2
z1 r1
  cos(1   2 )  i sin( 1   2 )
z 2 r2
And, similarly as before, we observe an r(cos + i sin) format that z1/z2 has modulus r = r1/r2 and an
argument  = 1 – 2 = arg(z1) – arg(z2).
Therefore,
arg(z1/z2) = arg(z1) – arg(z2).
…..shown.
■
(c) Starting with the standard inequality result |z1 + z2|  |z1| + |z2|, prove that |z1 – z2|  |z1| – |z2|.
 Hint: Replace z1 by (z1 – z2) without loss of generality.

|z1 + z2|  |z1| + |z2|
Substituting (as suggested), the complex number z1 with (z1 – z2) in the triangular inequality gives:
|(z1 – z2) + z2|  |(z1 – z2)| + |z2|

 |z1 – z2| + |z2|
|z1|
|z1| – |z2|  |z1 – z2|
And, reading the above inequality in the opposite direction (right-to-left):

|z1 – z2|  |z1| – |z2|
…..shown.
■
2. By considering the individual modulus and argument of each bracketed term, use the rules of complex
numbers (products, quotients, and powers) to express the following complex numbers in polar form.
(a)
 1  i 3 1  i 
 3  i
 We can denote the above complex number by z, and its modulus and argument by r and ,
respectively. Then applying appropriate rules means:
y
r
1 i 3 1 i
3 i
1 3 11

3 1
 2
3
1



  arg  1  i 3  arg 1  i   arg 3  i
  [180  60 ]  [45 ]
o
o
o

-1
 [360  30 ]
o
o
  120 o  45o  330 o
60o
O
45o
30o 1
3
x
-1
  120 o  45o  330 o
 3
  165 o
   360 165 o  195 o (using a positiveangle,anti - clockwiseabout O.)
In radians: 195 = 180 + 15 = 180 + ½ (30)   + ½(/6) =  + /12 = 13/12 rad.
Hence:
=
13
.
12
 Required polar form is:
13
13 

2  cos   i sin   .
12
12 

■
2
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
 1  i 3  1  i 
 3  i
3
(b)
Group I: CIV, ELEC, SURV.
4
5
 Again, denoting the modulus and argument by r and , then:
r
3
1 i 3
3i
1 i
3

2  2 

25
4
4
5
 2  2 4   1


  
  3 arg  1  i 3   4 arg 1  i   5 arg 

3  i
3
5
  arg   1  i 3   arg 1  i 4  arg  3  i 




  3 (120 o )  4 (45o )  5 (30 o )
  30 o
In radians:  = /6



 Required polar form is:  cos  i sin  .
6
6

1  i 3   1  i 
5
(c)
■
6
 2  2i 4
 Denoting the modulus and argument by r and , then:
1  i 3   1  i 
5
r

1
2
6
 2  2i 4
 1 i 3 5 1 i 6


4
 21  i 


 
 
  2   2  


 25 

 24 

1
2




1
2
2 
4
2 
6
2
1
2
 2  2  2
1
 r 2


5
 1  i 3  1  i 6
 arg 

 2  2i 4













 

 
 

1
2
 1  i 3 5  1  i 6 
1

arg 
4

2 



2

2
i


5
1
6
4
arg 1  i 3  arg  1  i   arg  2  2i  


2
1
5 arg 1  i 3  6 arg  1  i   4 arg  2  2i 
2
1
5 60 o  6 (135 o )  4 225 o
2


  




3
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
  = 105o
In radians:  = 105 × /180 = 7/12
7
7 

 Required polar form is: 2 cos
 i sin
.
12
12 

■
3. (a) Show that the locus of a complex number z = x + yi moving so that it is equidistant from two fixed
points represented by the complex numbers z1 = a + bi and z2 = c + di is a straight line that is the
perpendicular bisector of the line segment connecting z1 and z2 .
 Distance of z from point z1 equals distance of z from point z2. Mathematically, in complex number
language, this means:
z  z1  z  z 2

( x  iy )  (a  bi)  ( x  iy )  (c  di)
( x  a)  i( y  b)  ( x  c)  i( y  d )
Squaring both sides:
2
2
( x  a )  i ( y  b)  ( x  c )  i ( y  d )

( x  a) 2  ( y  b) 2  ( x  c) 2  ( y  d ) 2
Expanding both sides:
x 2  2ax  a 2  y 2  2by  b 2  x 2  2cx  c 2  y 2  2dy  d 2
The x2 and y2 terms cancel to give:
 2ax  a 2  2by  b 2  2cx  c 2  2dy  d 2
Keeping the y-terms on the L.H.S. :
2dy  2by  2ax  2cx  c 2  d 2  a 2  b 2
Factorizing:
2(d  b) y  2(a  c) x  c 2  d 2  a 2  b 2
And, solving for y explicitly:
c2  d 2  a2  b2
( a  c)
y
x
(d  b)
2(d  b)
The above equation for the locus of z is the standard form “y = mx + c”, and shows that the locus of z is
a straight line. Let’s call this line L1, its gradient being given by: m1 = (a – c)/(d – b).
We now test for L1 being perpendicular to the line segment (L2) connecting points z1(a, b) and z2(c, d).
The gradient (m2) of line segment L2 is calculated by the customary m 
m2 
y 2  y1
formula, so that:
x 2  x1
d b
ca
(a  c) d  b (a  c)
(a  c)



 1 .
( d  b) c  a (c  a )  ( a  c )
With the product m1m2 being equal to –1, then the locus of z (line L1) is indeed perpendicular to line
segment L2 .
To finally confirm the status of L1 being a perpendicular bisector of L2, we now have to show that L1
passes through the midpoint of line segment L2.
Product of gradients of lines L1 and L2 gives: m1 m2 
Recall that the coordinates of the midpoint (M) of any the line segment is given generally by :
 x  x 2 y1  y 2 
M 1
,

2 
 2
So for the line segment L2 connecting the points (a, b) and (c, d), its midpoint (M) has coordinates:
ac bd 
M
,
.
2 
 2
4
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
To check if L1 passes through M, we simply substitute the x-coordinate of M (that is: x = (a + c)/2) into
the ‘locus of z’ equation and see if we get the corresponding y-coordinate of M (that is: y = (b + d)/2).
c2  d 2  a2  b2
( a  c)
x
(d  b)
2(d  b)
Substituting x = (a + c)/2 into the above equation gives:
c2  d 2  a2  b2
( a  c) ( a  c)
y


(d  b)
2
2(d  b)
2
2
c  d  a2  b2
(a  c)(a  c)
y

2(d  b)
2(d  b)
Recall, locus of z (which is the equation of L1) : y 
y
y
y
y
y

y
c2  d 2  a2  b2
a2  c2

2(d  b)
2(d  b)
a2  c2  c2  d 2  a2  b2
2(d  b)
2
2
d b
2(d  b)
(d  b)( d  b)
2(d  b)
( d  b)
2
bd
……. same y-coordinate of midpoint M.
2
So we have confirmed that, indeed, the locus of z, is the perpendicular bisector of the line segment
connecting the complex numbers z1 and z2 .
■
(d) Find the locus of the complex number z defined by |2z + 1| = |z – 1|. Describe fully its geometry.
 Let z = x + iy, then |2z + 1| = |z – 1| gives:
|2(x + iy) + 1| = |(x + iy) – 1|
|(2x + 1) + i(2y)| = |(x – 1) + iy|
Squaring both sides:
|(2x + 1) + i(2y)|2 = |(x – 1) + iy|2
(2x + 1)2 + (2y)2 = (x – 1)2 + y2
Expanding both sides:
4x2 + 4x + 1 + 4y2 = x2 – 2x + 1 + y2
3x2 + 6x + 3y2 = 0

x2 + 2x + y2 = 0
Ans.
Now for a full geometrical description of the above locus equation:
Completing the square gives:
(x2 + 2x + 1) + y2 = 1

(x + 1)2 + y2 = 1
And the above equation represents a circle with its centre at point (–1, 0), and having a radius of
1 unit.
■
(e) Show that the locus of a complex number z defined by |z – 1| + |z + 1| = 4 is an ellipse having
equation:
y2
x2

 1 . Express in words, the motion of z based on the initial modulus-based equation given.
4
3
 We are given: |z – 1| + |z + 1| = 4
And we plan to square each side to break down the items, but this would be complicated if the
equation is left in its present term-arrangement. So we put the |z + 1| term on the R.H.S. to get:
5
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
|z – 1| = 4 – |z + 1|
Now squaring both sides:
|z – 1|2 = (4 – |z + 1| )2
|z – 1|2 = 16 – 8|z + 1| + |z + 1|2
Now substituting z = x + iy into the squared terms gives:
|(x + iy) – 1|2 = 16 – 8|z + 1| + |(x + iy) + 1|2
(x – l)2 + y2 = 16 – 8|z + 1| + (x + 1)2 + y2
Expanding:
x2 – 2x + 1 + y2 = 16 – 8|z + 1| + x2 + 2x + 1 + y2
And the x2 and y2 terms cancel, therefore:
– 2x = 16 – 8|z + 1| + 2x

8|z + 1| = 4x + 16
2|z + 1| = x + 4
Squaring both sides:
4|z + 1|2 = (x + 4)2
4|(x + 1) + iy|2 = (x + 4)2

4 [(x + 1)2 + y2] = (x + 4)2
Expanding:
4[x2 + 2x + 1 + y2] = x2 + 8x + 16
4x2 + 8x + 4 + 4y2 = x2 + 8x + 16

3x2 + 4y2 = 12
And dividing through by 12 gives:
x2 y2

1
4
3
…..shown.
■
Now for the motion depicted by the initial modulus-based equation:
We were given: |z – 1| + |z + 1| = 4.
This equation [geometrically] means that the complex number z moves so that the sum of its distances
from two fixed points (1, 0) and (0, –1) is a constant (equal to 4 units).
 zi  π
  . Describe fully its geometry.
z  2 4
(f) Find the equation of the locus defined by : arg
 Starting with
 zi  π
arg 

z  2 4
, then breaking down the quotient gives:
arg(z  i)  arg(z  2) 

π
4
arg(x  iy  i)  arg(x  iy  2) 
π
4
arg[x  i(y 1)]  arg[(x  2)  iy] 
π
4
y  
 y 1
1 
tan 1 
  tan 

 x 
 x  2 4
Taking the tangent of both sides, using the tan(A – B) formula on the L.H.S., then:
y 1
y


x
x2
 tan
4
 y  1  y 
1 


 x  x  2 
6
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
y 1
y

x
x2
 1
 y  1  y 
1 


 x  x  2 
Multiplying the numerator and denominator by x(x + 2) gives:
( y  1)( x  2)  ( y )( x)
1
x( x  2)  ( y  1) y
Expanding:
yx  2 y  x  2  yx
1
x 2  2x  y 2  y
2y  x  2
1
x  2x  y 2  y
2
Cross multiply:
2 y  x  2  x 2  2x  y 2  y
0  x 2  2x  y 2  y  2 y  x  2
0  x 2  3x  y 2  3 y  2
 x 2  3x  y 2  3 y  2  0
Ans.
Now for a full geometric description, we complete the square as follows:
x 2  3x  y 2  3 y  2
x
2
 
 3x   32   y 2  3 y   32 
2
 x  32   y  32   52
2
2
  2 
3 2
2
  32 
2
2
And the above equation defines a circle having centre  32 , 32  and radius of
5
2
units. ■
4. Obtain the following transformations (T) results from the xy-plane onto the uv-plane, where z = x + iy
and w = u + iv:
(a) Map the straight line y = 2x + 1 under T: w = 2z – 1, to get the image v = 2u + 4 .
 w = 2z – 1, implies:
u + iv = 2(x + iy) – 1
u + iv = (2x – 1)+ i(2y)
Therefore:
u = 2x – 1 and v = 2y
u 1
v
and y 
 x
2
2
But recall y = 2x + 1, hence:
v
 u 1
 2
 1
2
 2 
v
u2
2
 v  2u  4
■
1
z
(b) Map the circle x2 + y2 = 1 under T: w  z  , to get the image v = 0 (a horizontal line through O).
 wz
1
z
implies:
u  iv  x  iy 
1
x  iy
And we should by now know the result of the rationalizing of
1
x  iy
to be equal to
x  iy
x2  y2
.
7
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
Hence:
u  iv  x  iy 
x  iy
x2  y2
u  iv  x  iy 
 y 
x

 i 2
2
2 
x y
x y 
2

x  
y 
  i y  2

u  iv   x  2
2 
2 
x

y
x

y

 



x 

u   x  2
x  y 2 

and

y 

v   y  2
x  y 2 

But recall that x2 + y2 = 1 (the given z-plane geometry), hence we now have:

x

u  x 
1

and
y

v y 
1

u = 2x
and
v=0
So the image of the transformation will always lie on the horizontal line v = 0.
■
1
z
(c) Map the straight line y = 2x – 1 under T: w  , to get the image u2 – 2u + v2 – v = 0 (a circle).

w
1
z
implies:
u  iv 
1
x  iy
u  iv 
x  iy
x2  y2

u  iv 
 y 
x

 i 2
2
2 
x y
x y 

u
2
x
x  y2
2
and
v
y
x  y2
2
Getting rid of the complicated x2 + y2 items above as follows:
u x

v
y
Now recall that y = 2x – 1 (the given z-plane geometry), hence:
u
x

v 2x  1
or:
u
x

v 1  2x
And solving for x:
u (1  2 x)  vx
u  2ux  vx
u  2ux  vx
u  (2u  v) x

x
u
2u  v
 u 
 y  2 x  1  y  2
 1
 2u  v 
v
 2u   2u  v 
 y

 
 2u  v   2u  v  2u  v
8
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
 Collectively at this stage we have: x 
Group I: CIV, ELEC, SURV.
u
v
and y 
.
2u  v
2u  v
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
We now substitute these ‘values’ of x and y into either one of the u- and v- equations obtained
previously.
x
y
That is, into either u  2
.
or v  2
2
x y
x  y2
Let’s use the u-equation, to get:
u (2u  v)
u
2
u
v2

(2u  v) 2 (2u  v) 2

u
u
(2u  v) 2
 2
2u  v u  v 2
u (2u  v)
u2  v2
(2u  v)
1 2
u  v2
u 2  v 2  2u  v
u



u 2  v 2  2u  v  0 ■
5. By substituting  = ix, into the following trigonometric identities, derive their analogous hyperbolic
forms:
(a) cos2 + sin2 = 1


cos2(ix) + sin2(ix) = 1
(cos ix)2 + (sin ix)2 = 1
(cosh x)2 + (i sinh x)2 = 1
cosh2x – sinh2x = 1
Ans.
(b) sec2 = 1 + tan2

sec2(ix) = 1 + tan2(ix)
1
 1  (tan ix ) 2
2
(cos ix )
1
 1  (i tanh x) 2
(cosh x) 2

sech2x = 1 – tanh2x
Ans.
(c) sin2 = 2 sin cos


sin2(ix) = 2 sin(ix) cos(ix)
sin i(2x) = 2 (i sinh x) (cosh x)
i sinh 2x = i [2 sinh x cosh x]
sinh 2x = 2 sinh x cosh x
Ans.
(d) cos2 = cos2 – sin2 


cos2(ix) = cos2(ix) + sin2(ix)
cos i(2x) = (cos ix)2 + (sin ix)2
cosh 2x = (cosh x)2 + (i sinh x)2
cosh 2x = cosh2x – sinh2x
Ans.
9
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
e ix  e ix
e ix  e ix
and cos x 
to show that: sin3 cos2 = ½ (sin5 + sin).
2i
2
e i 3  e i 3 e i 2  e i 2
sin 3 cos 2 

2i
2
6. (a) Use the results sin x 

e i 5  e i  e i  e i 5
4i
Grouping according to the standard results given at the beginning of the question:
(e i 5  e i 5 )  (e i  e i )
sin 3 cos 2 
4i
(2i sin 5 )  (2i sin  )
sin 3 cos 2 
4i
sin 3 cos 2 
2i (sin 5  sin  )
4i
1
sin 3 cos 2  2 (sin 5  sin  )
sin 3 cos 2 

…..shown.
■
(b) Use the same technique in part (a) to obtain the more general identity:
sinA cosB = ½ [sin(A+B) + sin(A – B)].

sin A cos B 
e iA  e iA e iB  e iB

2i
2
e i ( A  B )  e i ( A B )  e i (  A  B )  e  i ( A  B )
sin A cos B 
4i
Working with (A + B) and (A – B) as the exponential-angles gives:
e i ( A  B )  e i ( A B )  e  i ( A  B )  e  i ( A  B )
sin A cos B 
4i
And then grouping appropriately:
(e i ( A  B )  e i ( A  B ) )  (e i ( A  B )  e  i ( A  B ) )
sin A cos B 
4i
2i sin( A  B)  2i sin( A  B)
sin A cos B 
4i
2i [sin( A  B)  sin( A  B)]
4i
1
sin A cos B  2 [sin( A  B)  sin( A  B)]
sin A cos B 

…..shown.
■
7. For z = x + iy, obtain the following in the form a + ib, hence calculate their moduli and arguments:
(i) (a) sin z
 sin z
= sin (x + iy)
= sin x cos iy + cos x sin iy
= sin x cosh y + cos x (i sinh y)
= (sin x cosh y) + i(cos x sinh y)
■
sin 2 x cosh 2 y  cos 2 x sinh 2 y ■
cos x sinh y
 tan 1 (cot x tanh y ) ■
arg(sin z) = tan 1
sin x cosh y
|sin z| =
(b) sin z
 sin z
= sin (x – iy)
= sin x cos iy – cos x sin iy
= sin x cosh y – cos x (i sinh y)
= (sin x cosh y) + i(–cos x sinh y)
■
10
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Group I: CIV, ELEC, SURV.
sin 2 x cosh 2 y  cos 2 x sinh 2 y ■
 cos x sinh y
arg(sin z) = tan 1
 tan 1 ( cot x tanh y )
sin x cosh y
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
|sin z| =
■
(c) tan z
 tan z
= tan (x + iy)
tan x  tan iy
=
1  tan x tan iy
tan x  i (tanh y )
=
1  i (tan x tanh y )
tan x  i (tanh y ) 1  i(tan x tanh y )
=

1  i (tan x tanh y ) 1  i(tan x tanh y )
=
[tan x  tan x tanh 2 y ]  i[tan 2 x tanh y  tanh y ]
1  tan 2 x tanh 2 y
=
tan x (1  tanh 2 y )  i tanh y (1  tan 2 x)
1  tan 2 x tanh 2 y
=
tan x sec h 2 y  i tanh y sec 2 x
1  tan 2 x tanh 2 y
|tan z| =
=
=
■
tan x sec h 2 y  i tanh y sec 2 x
….. the denominator is already positive & real.
1  tan 2 x tanh 2 y
(tan x sec h 2 y ) 2  (tanh y sec 2 x) 2
1  tan 2 x tanh 2 y
tan 2 x sec h 4 y  tanh 2 y sec 4 x
■
1  tan 2 x tanh 2 y
arg(tan z) = tan
1
tanh y sec 2 x
tan x sec h 2 y
.....{acceptable}
tanh y cosh 2 y
(cosh y tanh y ) cosh y
 tan 1
= tan
2
(cos x tan x) cos x
tan x cos x
sinh y cosh y
= tan 1
■
sin x cos x
1
(d) cos z
 cos z = cos (x + iy)
= cos x cos iy – sin x sin iy
= cos x cosh y – sin x (i sinh y)
= (cos x cosh y) + i(–sin x sinh y)
|cos z|
arg(cos z)
=
■
cos 2 x cosh 2 y  sin 2 x sinh 2 y
  sin x sinh y 

= tan 1 
cos
x
cosh
y


1
= tan ( tan x tanh y)
■
■
(e) cos z
 cos z = cos (x – iy)
= cos x cos iy + sin x sin iy
= cos x cosh y + sin x (i sinh y)
= (cos x cosh y) + i(sin x sinh y)
■
11
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Group I: CIV, ELEC, SURV.
cos 2 x cosh 2 y  sin 2 x sinh 2 y ■
sin x sinh y
arg(cos z ) = tan 1
 tan 1 (tan x tanh y )
cos x cosh y
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
|cos z | =
■
(f) cot z
 cot z =
1
tan z
1
tan( x  iy )
1  tan x tan iy
=
tan x  tan iy
1  i (tan x tanh y )
=
tan x  i (tanh y )
1  i (tan x tanh y ) tan x  i (tanh y )

=
tan x  i (tanh y ) tan x  i (tanh y )
=
[tan x  tan x tanh 2 y ]  i [ tanh y  tan 2 x tanh y ]
=
tan 2 x  tanh 2 y
tan x (1  tanh 2 y )  i [(  tanh y )(1  tan 2 x)]
=
tan 2 x  tanh 2 y
tan x sec h 2 y  i ( tanh y sec 2 x)
=
tan 2 x  tanh 2 y
■
And so:
|cot z| =
=
tan x sec h 2 y  i ( tanh y sec 2 x)
tan 2 x  tanh 2 y
tan 2 x sec h 4 y  tanh 2 y sec 4 x)
■
tan 2 x  tanh 2 y
  tanh y sec 2 x 

arg(cot z) = tan 
2
tan
x
sec
h
y


2
  tanh y cosh y
= tan 1 
2
 tan x cos x
1
……{acceptable, may stop here}

 (cosh y tanh y) cosh y 
  tan 1 

 (cos x tan x) cos x 

  sinh y cosh y 

= tan 1 
sin
x
cos
x


■
(ii) It is seen that sin z = sin z , and cos z = cos z , use this fact to prove that: tan z = tan z .
 Hint: Recall that the quotient of the conjugates equals the conjugate of the quotient.
 tan z 
sin z
sin z
 sin z 


  tan z
cos z
cos z  cos z 
…..shown.
■
  sinh 2 y 
.
 sin 2 x 
(iii) Also show that: argcot z   tan 1 
 The number of steps necessary will depend on how far the expression for arg(cot z) was reduced in
part i(f) above.
In this context, we had obtained:
  sinh y cosh y 

arg(cot z) = tan 1 
 sin x cos x 
Therefore, creating the required identities inside the R.H.S. brackets, returns:
12
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #1: COMPLEX NUMBERS I
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
  2 sinh y cosh y 

arg(cot z) = tan 1 
2
sin
x
cos
x



  sinh 2 y 
arg(cot z) = tan 1 

 sin 2 x 
…..shown.
■
E N D O F S O L U T I O N S.
13