Understanding Engineering Mathematics

which we can check directly

√

3 −j

1 −

√

3 j

=

(

√

3 −j)( 1 +

√

3 j)

12 +(

√

3 )^2

=

2

√

3 + 2 j

4

=

√

3

2

+

1

2

j

Exercise on 12.5

Ifz 1 =^

(π

4

)

andz 2 =^

(

−

π

3

)

evaluate

z 1

z 2

in polar and Cartesian form.

Answer
z 1
z 2

=^

(

7 π

12

)

=

( 1 −

√

3 )

2

√

2

+

(

√

3 + 1 )

2

√

2

j

12.6 Exponential form of a complex number

Having explained the usefulness of the polar form in multiplication and division of complex
numbers, I will now introduce a result that is as pretty as it is powerful:

ejθ≡cosθ+jsinθ=^ (θ)

Remember that this result is unchanged ifθis replaced byθ+ 2 kπwherekis an integer.
This result forejθwas first stated explicitly by the Swiss mathematician Leonard Euler
(1707 – 83) in 1748 – not so long ago really. It is known, along with countless other results,
asEuler’s formula. It can be proved by expanding the left-hand side in series, gathering
together real and imaginary parts after usingj^2 =−1 and summing the resulting series to
cosθand sinθ(see Reinforcement Exercise 18).

Problem 12.10

By taking particular values forqshow that

(i)ej^0 =ej^2 p= 1 (ii)ejp=− 1 (iii)ej

p

(^2) =j
(i)ej^0 =cos 0+jsin 0=1 and similarlyej^2 π=cos 2π+jsin 2π= 1
(ii)ejπ=cosπ+jsinπ=− 1
(iii)ej
π
(^2) =cos
π
2
+jsin
π
2
= 0 +j 1 =j
It can be shown thatejθobeys all the usual indices laws:
ejxejy=ej(x+y)– from multiplication in polar form
ejx
ejy
=ej(x−y)– from division in polar form
(ejx)y=ejxy– from an extension of Q2, Exercise 12.4