Solution to EE 503 Test #1 S03 (Version 1)
(Solutions to other versions are similar.)
1.
Find the numerical values of the magnitude and angle of any two different complex
numbers, z1 and z2 which both satisfy the equation, exp( z) = 1 − j . That is, exp( z1 ) = 1 − j and
exp( z2 ) = 1 − j .
Magnitude of z1 = 0.8585
Angle (Phase) of z1 =
-1.1552
Magnitude of z2 = 5.5087
or
Magnitude of z2 = 7.0771
or
.
.
.
(an infinity of solutions)
Angle (Phase) of z2 =
-1.5078
Angle (Phase) of z2 =
-1.5218
π
−j 

 π

z = log(1 − j ) = log 2e 4  = ln 2 + j  − + 2 nπ 
 4



( )
( )
π
= 0.3466 − j 0.7854 = 0.8585e − j1.1552
4
( )
7π
= 0.3466 + j 5.4978 = 5.5087e j1.5078
4
( )
9π
= 0.3466 − j 7.0686 = 7.0771e − j1.5218
4
z1 = ln 2 − j
z2 = ln 2 + j
or
z2 = ln 2 − j
Alternate Solution:
exp( z) = e x (cos( y ) + j sin( y )) = 1 − j
e x cos( y ) + je x sin( y ) = 1 − j
e x cos( y ) = 1 , e x sin( y ) = −1
Forming the ratio of these two equations,
sin( y )
π
= tan( y ) = −1 ⇒ y = tan −1 (−1) = − + 2 nπ
cos( y )
4
(Note: y ≠
3π
+ 2 nπ because e x cos( y ) = 1 and e x sin( y ) = −1.)
4
 π

 π
Then e x cos −  = 1 ⇒ e x = 2 ⇒ x = ln 2 . Therefore z = ln 2 + j  − + 2 nπ  as in the
 4

 4
( )
first solution and the rest of the solution is the same.
( )
2.
If, on the branch, 0 ≤ θ < 2π , we choose to find the square root of j2 as
j 2 = 2e
j
π
2
= 2e
j
π
4
= 1+ j,
20 z
, at the pole, 2 (1− j ) , on the
z 4 + 16
branch, 0 ≤ θ < 2π , finding square roots the same way (magnitude of the square root is the
positive square root of the magnitude and the angle of the square root is half the angle). Report
the numerical value in either of the forms, x + jy or re jθ .
find the numerical value of the residue of the function,
20 z
20 z
=
4
π
3π
5π
7π
j 
j
j
j



z + 16 
4
4
4
4
 z − 2e   z − 2e   z − 2e   z − 2e 





Residue at z = 2 (1 − j ) =
20 2e
j
7π
4
7π
3π
5π
π
j 
j
j
j
  j 74π

 j 74π
4
4
4
4
 2e − 2e   2e − 2e   2e − 2e 




Taking the square root according to the method prescribed,
Residue at z = 2 (1 − j ) =
j
7π
8
5 2
e
7π
7π
3π
5π
π
j  j
j
j
2  j4
  j 74π

4
4
4
4
e
−
e
e
−
e
e
−
e








j
7π
5 2
e 8
Residue at z = 2 (1 − j ) =
2 −j 2 2− j 2
(
j
)
7π
8
5 2e
5 2
Residue at z = 2 (1 − j ) = j
e
7π =
j
8
8
4
e
Residue at z = 2 (1 − j ) = 0.8839e
−j
j
3π
8
j
π
2
e
e
7π
j
7π
e 8
5 2 e 8
=−
7π
j
j4
2− j 2
2e 4
5 2
=−
j4
2
j
7π
8
j
7π
4
5 2 − j 38π
=
e
8
= 0.3382 − j 0.8166
or, expressing the result with an angle on the specified branch (which is not necessary)
Residue at z = 2 (1 − j ) = 0.8839e j 5.1051.
3.
Find the numerical value of the integral,
∞
I=∫
0
dx
,
x +1
4
by means of a contour integral in the complex plane using the contour illustrated below as
R → ∞.
y
[z]
R
x
Since the integrand is even,
∞
I=
dx
dz
dz
1
1
1
=
=
4
4
∫
∫
∫
−1 + j  
−1 − j 
2 −∞ x + 1 2 C ,ccw z + 1 2 C ,ccw  1 + j   1 − j  
z −
z −
z −
z −








2
2
2
2 
By Jordan’s lemma the integral along the semicircular contour is zero, therefore
I=
1
× j 2π ∑ residues
2
jπ



  1 + j 1 − j   1 + j −1 + j   1 + j −1 − j 
−
−
−





2  2
2  2
2 

 2
I=

jπ
+

  −1 + j − 1 + j   −1 + j − 1 − j   −1 + j − −1 − j  
  2
2  2
2  2
2  


1
1
I = jπ 2 2 
+

 j 2 × 2 × (2 + j 2) −2 × (−2 + j 2) × j 2 
I=
jπ 2 2  1
1  π 2 1
1 
−
+

=


j 4  2 + j 2 −2 + j 2 
2  2 + j 2 2 − j 2
I=
π 2  2 − j 2 + 2 + j 2 π 2

=

2 
8
4