Trigonometric form of a complex number, de Moivre's formula
Use the above figure to express the real and imaginary parts a,b
of the complex number in terms of the magnitude and argument ρ, j.
a=r.cosj, b=r.sinj.
Hence
a=r(cosj+isinj)
This is the trigonometric form of a complex number.
Each complex number has a unique trigonometric form because it has a unique magnitude and argument.
Example. α= .
; .
Therefore,
.
1. Multiplication. Consider complex numbers
a1 =r1 (cosj1 +isinj1) and a2 =r2 (cosj2 +isinj2).
The product equals
a1·a2 = r1 (cosj1 +isinj1). r2 (cosj2 +isinj2)=
=r1·r2 [(cosj1cosj2 -sinj1sinj2)+i(sinj1cosj2 + cosj1sinj2);
a1·a2 = r1·r2 [cos(j1+j2 )+isin(j1+j2 )].
Therefore, in order to multiply two complex numbers, it is necessary to multiply their magnitudes and add the arguments.
2.Division. Multiply both the numerator and denominator by the conjugate of the denominator
![](http://ok-t.ru/doclecturenet/baza1/1672321084415.files/image3148.gif)
Therefore, in order to divide two complex numbers, it is necessary to divide their magnitudes and subtract the arguments.
A power of a complex number. If a1=a2=a3=…=an=a, then
an = r ·r ·r·…·r[cos(j+j +j+…+j)+i(sin(j+j +j+…+j)]= =rn[cosnj+isinnj]
or
ak =rk[coskj+isinkj] - de Moivre's formula.
Roots of a complex number. Write down de Moivre's formula for using the fact the sine and cosine are periodic functions with the period T=2p:
,
where k may be any of n integers 0,1,2,…,n-1.
Example. Solve the equation x3+1=0.
1-st method: (x+1)(x2-x+1)=0 , x1=-1, ;
2-nd method: x3=-1, , -1=cosp+isinp, which is a trigonometric form of a number.
. Using the formula, k=0, ,
k=1, ,
k=2, ![](http://ok-t.ru/doclecturenet/baza1/1672321084415.files/image3164.gif)
= .
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Date: 2015-01-02; view: 1598
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