Understanding Engineering Mathematics

(やまだぃちぅ) #1
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
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