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
a_{1} =r_{1} (cosj_{1} +isinj_{1}) and a_{2} =r_{2} (cosj_{2} +isinj_{2}).
The product equals
a_{1}·a_{2} = r_{1} (cosj_{1} +isinj_{1})^{.} r_{2} (cosj_{2} +isinj_{2})=
=r_{1}·r_{2} [(cosj_{1}cosj_{2} sinj_{1}sinj_{2})+i(sinj_{1}cosj_{2} + cosj_{1}sinj_{2});
a_{1}·a_{2} = r_{1}·r_{2} [cos(j_{1}+j_{2} )+isin(j_{1}+j_{2} )].
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
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 a_{1}=a_{2}=a_{3}=…=a_{n}=a, then
a^{n} = r ·r ·r·…·r[cos(j+j +j+…+j)+i(sin(j+j +j+…+j)]= =r^{n}[cosnj+isinnj]
or
a^{k} =r^{k}[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,…,n1.
Example. Solve the equation x^{3}+1=0.
1st method: (x+1)(x^{2}x+1)=0 , x_{1}=1, ;
2nd method: x^{3}=1, , 1=cosp+isinp, which is a trigonometric form of a number.
. Using the formula, k=0, ,
k=1, ,
k=2,
= .
ELABORATIONS OF PRACTICAL CLASSES
Date: 20150102; view: 1487
