De Moivre’s theorem 229
Problem 6. Change (3 − j 4) into (a) polar form,
(b) exponential form.
(a) (3 − j 4) = 5∠−53.13
◦ or 5∠−0.927
in polar form
(b) (3 − j 4) = 5∠−0.927 = 5e −j0.927
in exponential form
Problem 7. Convert 7.2e j 1.5 into rectangular
form.
7.2e
j 1.5
= 7.2∠1.5 rad(= 7.2∠85.94
◦
) in polar form
= 7.2 cos 1.5 + j 7.2 sin1.5
= (0.509 + j 7.182) in rectangular form
Problem 8. Express z = 2e
1+ j
π
3 in Cartesian
form.
z = (2e
1
)
e
j
π
3
by the laws of indices
= (2e
1
)∠
π
3
(or 2e∠60
◦
)in polar form
= 2e
cos
π
3
+ j sin
π
3
= (2.718 + j4.708) in Cartesian form
Problem 9. Change 6e 2− j 3 into (a + j b) form.
6e
2− j 3
= (6e
2
)(e
− j 3
) by the laws of indices
= 6e
2
∠−3 rad (or 6e
2
∠−171.89
0
)
in polar form
= 6e
2 [cos (−3) + j sin (−3)]
= (−43.89 − j6.26) in (a + jb) form
Problem 10. If z = 4e j 1.3 , determine ln z (a) in
Cartesian form, and (b) in polar form.
If z = re
j θ then ln z = ln(re
j θ
)
= lnr + ln e
j θ
i.e.
ln z = lnr + j θ,
by the laws of logarithms
(a) Thus if z =4e
j 1.3 then ln z = ln(4e
j1.3
)
= ln 4 + j1.3
(or 1.386 + j1.300) in Cartesian form.
(b) (1.386 + j 1.300) =1.90∠43.17 ◦ or 1.90∠0.753
in polar form.
Problem 11. Given z = 3e 1− j , find ln z in polar
form.
If
z = 3e
1− j
, then
ln
z = ln(3e
1− j
)
= ln 3 + ln e
1− j
= ln 3 + 1 − j
= (1 + ln 3) − j
= 2.0986 − j 1.0000
= 2.325∠−25.48
◦ or 2.325∠−0.445
Problem 12. Determine, in polar form, ln (3 + j 4).
ln(3 + j 4) = ln[5∠0.927] = ln[5e
j 0.927 ]
= ln 5 + ln(e
j 0.927
)
= ln 5 + j 0.927
= 1.609 + j 0.927
= 1.857∠29.95
◦ or 1.857∠0.523
Now try the following exercise
Exercise 92 Further problems on the
exponential form of complex numbers
1. Change (5 + j 3) into exponential form.
[5.83e j 0.54 ]
2. Convert (−2.5 + j 4.2) into exponential form.
[4.89e j 2.11 ]
3. Change 3.6e j 2 into cartesian form.
[−1.50 + j 3.27]
4. Express 2e
3+ j
π
6 in (a + j b) form.
[34.79 + j 20.09]
5. Convert 1.7e
1.2− j 2.5 into rectangular form.
[−4.52 − j 3.38]