Understanding Engineering Mathematics

Problem 12.11

Put into exponential form (i) 1 (ii)−1 (iii)− 2 j (iv) 1Y

√

3 j

(v)

√

3 −j

Conversion to exponential form is of course no more than converting to polar form – i.e.
findingrandθ.

(i) z=1, r=1, θ=0soz= 1 ej^0 =ej^0

(ii) z=−1, r=1, θ=π so z=ejπ

(iii) z=− 2 jr=2, θ=−

π

2

so z= 2 e−jπ/^2

(iv) z= 1 +

√

3 jr=2, θ=

π

3

so z= 2 ejπ/^3

(v) z=

√

3 −jr=2, θ=−

π

6

so z= 2 e−jπ/^6

Exercises on 12.6

1. Show that the modulus ofejθis one


2. Show that(ejθ)−^1 =e−jθ


3. Show that cosθ=

ejθ+ejθ

2

sinθ=

ejθ−e−jθ

2 j

12.7 De Moivre’s theorem for integer powers

A remarkable result ‘follows’ from the above Euler formula, if we assume that the usual
rules of indices apply. Ifnis an integer then

(cosθ+jsinθ)n=(ejθ)n=ejnθ=cosnθ+jsinnθ

giving usDe Moivre’s theorem for a positive integer:

Ifnis an integer, then

(cosθ+jsinθ)n=cosnθ+jsinnθ

Basically, this result is a generalization of the ‘add arguments to form product’ rule. As an
example we have(cosθ+jsinθ)^2 =cos(θ+θ)+jsin(θ+θ)=cos 2θ+jsin 2θ.And
if you were misguided enough to repeat this multiplication a few times you would find,
for example, that(cosθ+jsinθ)^5 =cos 5θ+jsin 5θ. De Moivre’s theorem gives this
to us directly however.

Problem 12.12

Using De Moivre’s theorem show that.

√

3 Yj/−^3 =−

1

8

j.

First note that we can write

√

3 +j= 2

(√

3

2

+

1 j

2

)

= 2

(

cos

π

6

+jsin

π

6

)

and so