Higher Engineering Mathematics

DE MOIVRE’S THEOREM 265

E

Problem 6. Change (3−j4) into (a) polar

form, (b) exponential form.

(a) (3−j4)= 5 ∠− 53. 13 ◦or 5 ∠− 0. 927

in polar form

(b) (3−j4)= 5 ∠− 0. 927 =5e−j^0.^927

in exponential form

Problem 7. Convert 7.2ej^1.^5 into rectangular

form.

7 .2ej^1.^5 = 7. 2 ∠ 1 .5 rad(= 7. 2 ∠ 85. 94 ◦) in polar form

= 7 .2 cos 1. 5 +j 7 .2 sin 1. 5

=(0.509+j 7 .182)in rectangular form

Problem 8. Expressz=2e^1 +j

π

(^3) in Cartesian
form.
z=(2e^1 )
(
ej
π
3
)
by the laws of indices
=(2e^1 )∠
π
3
(or 2e∠ 60 ◦)in polar form
=2e
(
cos
π
3
+jsin
π
3
)
=(2. 718 +j 4 .708)in Cartesian form
Problem 9. Change 6e^2 −j^3 into (a+jb) form.
6e^2 −j^3 =(6e^2 )(e−j^3 ) by the laws of indices
=6e^2 ∠−3 rad (or 6e^2 ∠− 171. 890 )
in polar form
=6e^2 [cos(−3)+jsin (−3)]
=(− 43. 89 −j 6 .26)in (a+jb) form
Problem 10. Ifz=4ej^1.^3 , determine lnz(a) in
Cartesian form, and (b) in polar form.
If z=rejθthen lnz=ln (rejθ)
=lnr+ln ejθ
i.e. lnz=lnr+jθ,
by the laws of logarithms
(a) Thus ifz=4ej^1.^3 then lnz=ln (4ej^1.^3 )
=ln 4+j 1. 3
(or 1. 386 +j 1. 300 ) in Cartesian form.
(b) (1. 386 +j 1 .300)= 1. 90 ∠ 43. 17 ◦or 1. 90 ∠ 0. 753
in polar form.
Problem 11. Givenz=3e^1 −j, find lnzin 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+j4).
ln(3+j4)=ln[5∠ 0 .927]=ln[5ej^0.^927 ]
=ln 5+ln(ej^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 107 Further problems on the
exponential form of complex numbers

1. Change (5+j3) into exponential form.
   [5.83ej^0.^54 ]


2. Convert (− 2. 5 +j 4 .2) into exponential form.
   [4.89ej^2.^11 ]


3. Change 3.6ej^2 into cartesian form.
   [− 1. 50 +j 3 .27]