Complex numbers 219
(iii) θ is called the argument (or amplitude) of Z and
is written as arg Z .
By trigonometry on triangle OAZ,
arg Z = θ = tan
−1 y
x
(iv) Whenever changing from cartesian form to polar
form, or vice-versa, a sketch is invaluable for
determining the quadrant in which the complex
number occurs.
Problem 9. Determine the modulus and argument
of the complex number Z = 2 + j 3, and express Z
in polar form.
Z = 2 + j 3 lies in the first quadrant as shown in
Fig. 20.5.
Real axis
2
Imaginary
axis
r
0
j3
Figure 20.5
Modulus, |Z | =r =
(2 2 + 3 2 ) =
√
13 or 3.606, correct
to 3 decimal places.
Argument, arg Z = θ = tan −1 3
2
= 56.31 ◦ or 56 ◦ 19
In polar form, 2 + j 3 is written as 3.606∠56.31
◦ .
Problem 10. Express the following complex
numbers in polar form:
(a) 3 + j 4
(b)−3 + j 4
(c) −3 − j 4 (d) 3 − j 4
(a) 3 + j 4 is shown in Fig. 20.6 and lies in the first
quadrant.
Modulus, r =
(3 2 + 4 2 ) = 5 and argument
θ = tan −1 4
3 = 53.13 ◦ .
Hence 3 + j4 = 5∠53.13 ◦
2
21
22
2j
j
2j2
j2
2j3
j3
2j4
Real axis
Imaginary
axis
␣
␣
␣
r
r
r
r
1
23
j4
(23 2 j4)
(23 1j4)
(3 1j4)
(3 2 j4)
3
Figure 20.6
(b) −3 + j 4 is shown in Fig. 20.6 and lies in the
second quadrant.
Modulus, r = 5 and angle α = 53.13 ◦ , from
part (a).
Argument =180
◦
− 53.13
◦
= 126.87
◦ (i.e. the
argument must be measured from the positive real
axis).
Hence −3 + j4 = 5∠126.87 ◦
(c) −3 − j 4 is shown in Fig. 20.6 and lies in the third
quadrant.
Modulus, r = 5 and α = 53.13 ◦ , as above.
Hence the argument = 180 ◦ + 53.13 ◦ = 233.13 ◦ ,
which is the same as −126.87 ◦ .
Hence (−3 − j4) = 5∠233.13 ◦ or 5∠−126.87 ◦
(By convention the principal value is normally
used, i.e. the numerically least value, such that
−π <θ <π).
(d) 3 − j 4 is shown in Fig. 20.6 and lies in the fourth
quadrant.
Modulus, r = 5 and angle α = 53.13 ◦ , as above.
Hence (3 − j4) = 5∠−53.13 ◦
Problem 11. Convert (a) 4∠30 ◦ (b) 7∠−145 ◦
into a + j b form, correct to 4 significant figures.
(a) 4∠30 ◦ is shown in Fig. 20.7(a) and lies in the first
quadrant.
(iii) θ is called the argument (or amplitude) of Z and
is written as arg Z .
By trigonometry on triangle OAZ,
arg Z = θ = tan
−1 y
x
(iv) Whenever changing from cartesian form to polar
form, or vice-versa, a sketch is invaluable for
determining the quadrant in which the complex
number occurs.
Problem 9. Determine the modulus and argument
of the complex number Z = 2 + j 3, and express Z
in polar form.
Z = 2 + j 3 lies in the first quadrant as shown in
Fig. 20.5.
Real axis
2
Imaginary
axis
r
0
j3
Figure 20.5
Modulus, |Z | =r =
(2 2 + 3 2 ) =
√
13 or 3.606, correct
to 3 decimal places.
Argument, arg Z = θ = tan −1 3
2
= 56.31 ◦ or 56 ◦ 19
In polar form, 2 + j 3 is written as 3.606∠56.31
◦ .
Problem 10. Express the following complex
numbers in polar form:
(a) 3 + j 4
(b)−3 + j 4
(c) −3 − j 4 (d) 3 − j 4
(a) 3 + j 4 is shown in Fig. 20.6 and lies in the first
quadrant.
Modulus, r =
(3 2 + 4 2 ) = 5 and argument
θ = tan −1 4
3 = 53.13 ◦ .
Hence 3 + j4 = 5∠53.13 ◦
2
21
22
2j
j
2j2
j2
2j3
j3
2j4
Real axis
Imaginary
axis
␣
␣
␣
r
r
r
r
1
23
j4
(23 2 j4)
(23 1j4)
(3 1j4)
(3 2 j4)
3
Figure 20.6
(b) −3 + j 4 is shown in Fig. 20.6 and lies in the
second quadrant.
Modulus, r = 5 and angle α = 53.13 ◦ , from
part (a).
Argument =180
◦
− 53.13
◦
= 126.87
◦ (i.e. the
argument must be measured from the positive real
axis).
Hence −3 + j4 = 5∠126.87 ◦
(c) −3 − j 4 is shown in Fig. 20.6 and lies in the third
quadrant.
Modulus, r = 5 and α = 53.13 ◦ , as above.
Hence the argument = 180 ◦ + 53.13 ◦ = 233.13 ◦ ,
which is the same as −126.87 ◦ .
Hence (−3 − j4) = 5∠233.13 ◦ or 5∠−126.87 ◦
(By convention the principal value is normally
used, i.e. the numerically least value, such that
−π <θ <π).
(d) 3 − j 4 is shown in Fig. 20.6 and lies in the fourth
quadrant.
Modulus, r = 5 and angle α = 53.13 ◦ , as above.
Hence (3 − j4) = 5∠−53.13 ◦
Problem 11. Convert (a) 4∠30 ◦ (b) 7∠−145 ◦
into a + j b form, correct to 4 significant figures.
(a) 4∠30 ◦ is shown in Fig. 20.7(a) and lies in the first
quadrant.