Assigned 8/31/01
EE301
Due 9/12/01
Fall 2001
Prof. Fowler
EE 301
HW#1
Complex Numbers & Complex Functions
NOTE: Show Steps Used
– don’t just use your calculator’s complex number functions!!
1.
For each of the following complex numbers, show its graphical representation on the
complex plane and convert it into polar form.
a.
2–
j
3
b.
6
c.
2
j
d.
2+
j
4
e.
–3–
j
4
f.
–4+
j
4
2.
For each of the following complex numbers, show its graphical representation on the
complex plane and convert it into rectangular form (show steps used – don’t just use
your calculator’s polarto rectangular function).
a.
3
e
j
π
/8
b.
4
e
j
π
c.
3
e
j
2
π
/3
d.
3
e
j
0
e.
3
e
j
π
/2
f.
3
e
j
4
π
3.
Find the following products
a.
(2–
j
3) 3
e
j
π
/8
b.
(2+
j
4) 3
e
j
2
π
/3
4.
Find the following quotients
a.
(2–
j
3)/3
e
j
π
/8
b.
(2+
j
4)/3
e
j
2
π
/3
5.
For each of the following complex numbers, simplify the result as appropriate (note:
in each case the number is rewritten in a form that is helpful to find the answer)
a.
je
j
π
/2
=
j
×
e
j
π
/2
b.
e
j
3
π
/4
= 1
×
e
j
3
π
/4
c.
4
/
4
/
π
π
j
j
e
j
je
×
−
=
−
6.
For each of the following complexvalued sequences of numbers, plot the magnitude
and angle versus the integer variable
}
,
3
,
2
,
1
,
0
,
1
,
2
,
3
,
{
…
…
−
−
−
∈
n
.
a.
z
(
n
) = 4 +
j
3
n
b.
z
(
n
) = (4 +
j
3
n
)/(4
n
+
j
3)
c.
z
(
n
) = e
j
4
π
n
7.
For each of the following functions, plot the magnitude and angle versus the real
variable
.
)
,
(
∞
−∞
∈
x
a.
z
(
x
) = 10/(4x +
j
3)
b.
z
(
x
) = e
j
4
π
x
c.
Plot the real and imaginary parts of the function in 7b
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EE 301
HW#1 Solutions
Note: ALWAYS state units of quantities
.
Here the magnitudes are unitless because we don’t
know what the complex numbers physically represent (i.e. voltage, etc.) BUT the angles are in
radians – so state their units as “rad”
1. Rect to Polar:
For each case you need to compute the magnitude and angle – sometimes it is
best to use the formulae but other times a bit of thinking provides an easy way to the answer
– always look for these things to save time on exams!!!
a.
2–
j
3
rad
983
.
0
)
2
/
3
(
tan
13
)
3
(
2


1
2
2
−
=
−
=
∠
=
−
+
=
−
z
z
983
.
0
13
z
j
e
−
=
⇒
Sanity Check: 0.983>
π
/2 so the angle is in 4
th
Quadrant as shown
b.
6
This one is easiest done by inspection of the graph:
6
0
6


=
⇒
=
∠
=
z
z
z
Note: for this one, polar = rect
Any positive real number has an angle of zero
Any negative real number has an angle of
±π
c.
2
j
This one is easiest done by inspection of the graph:
2
/
2
rad
2
/
2


π
=
⇒
π
=
∠
=
j
e
z
z
z
Any positive imaginary number has an angle of
π
/2
Any negative imaginary number has an angle of 
π
/2
d.
2+
j
4
107
.
1
1
2
2
20
rad
107
.
1
)
2
/
4
(
tan
20
4
2


j
e
z
z
z
=
⇒
=
=
∠
=
+
=
−
107
.
1
20
z
j
e
=
⇒
Sanity: 1.107 <
π
/2, which matches plot!!!
e.
–3–
j
4
rad
069
.
4
927
.
0
)
3
/
4
(
tan
5
25
)
4
(
)
3
(


1
2
2
=
π
+
=
−
−
=
∠
=
=
−
+
−
=
−
z
z
069
.
4
5
j
e
z
=
⇒
Note: the calculator gave the wrong quadrant so I had to
adjust by adding
π
.
You could also subtract
π
to get an
equivalent answer:
21
.
2
5
j
e
z
−
=
⇒
Re
Im
2
3
Re
Im
6
Re
Im
2
Re
Im
2
4
Re
Im
3
4
Calc
thinks
its here!
f.
–4+
j
4
2
4
16
2
)
4
(
)
4
(


2
2
=
×
=
+
−
=
z
Angle from inspection of graph:
4
/
3
π
=
∠
z
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 Fall '08
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 Complex number

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