Complex numbers:
In this chapter,we survey the algebraic and geometric structure of the complex numbers system,we assume various corresponding properties of real numbers to be known.
1-sums and products:
Complex numbers can be defined as order pairs (x,y) of real numbers coordinates x and y ,when real numbers x are displayed as points(x,0) on the real axis,it is clear that the set of complex numbers includes the real numbers as a subset,
Complex numbers of the form(0,y) correspond to points on the y axis are called pure imaginary numbers the y axis is then referred as the imaginary axis ,so that z=(x,y),the real numbers x and y ,moreover known as the real and imaginary parts of z,so that
Rez=x & Imz=y
Example:skerch or draw
Z1=(3,2),z2=(0,1),z3=(1,0)
Now ,two complex z1=(x1,y1)&z2=(x2,y2) are equal whenever the have the same real parts and the same imaginary parts,thus
z1=z2 ?Rez1=Rez2 & Imz1=Imz2
the sum z1+z2 and the product z1z2 of two complex numbers z1=(x1,y1),z2=(x2,y2) are defined as follows:
z1+z2=(x1,y1)+(x2,y2)=(x1+x2,y1+y2)
z1z2=(x1,y1).(x2,y2)=(x1x2-y1y2,y1x2+y2x1)
since
z1=(x1,y1)= x1+iy1
z2=(x2,y2)= x2+iy2
z1+z2=(x1+iy1)+(x2+iy2)=(x1+x2)+i(y1+y2)= (x1+x2,y1+y2)
and
z1z2=(x1+iy1)(x2+iy2)=x1x2+ix1y2+iy1x2+i2y1y2
=x1x2+ix1y2+iy1x2-y1y2
=(x1x2- y1y2)+i(y1x2+y2x1)
=(x1x2- y1y2,y1x2+y2x1)
Any complex number z=(x,y) can be written as z=(x,0)+(0,y)
Z=x+iy ,where i?pure imaginary number (0,1)
Also ,with the convention
Z2=zz ,z3=zzz,……etc, we find that
i2=(0,1)(0,1)=(-1,0)=-1
2-algebraic properties:
1-commutative law:
z1+z2=z2+z1 , z1z2=z2z1
2-associative law:
(z1+z2)+z3=z1+(z2+z3) , (z1z2)z3=z1(z2z3)
3-distributive law:
z(z1+z2)=zz1+zz2
according to the commutative law for multiplication iy=yi ,hence one can write z=x+yi instead of z=x+iy
the additive identity 0=(0,0) and the multiplication identity 1=(1,0) for real number carry over to the entire complex number system ,that is
z+0=z and z.1=z ? z-complex number
there is associated with each complex number z=(x,y) an additive inverse
-z=(-x,-y) ? z+(-z)=0
Since
(x,y)+(u,v)=(0,0)?u=-x ? v=-y
-z=-x-iy ? -(iy)=(-i)y=i(-y)
So ,if
z1=(x1,y1) ? z2=(x2,y2) ,then
z1-z2=(x1-x2,y1-y2)= (x1-x2)+i(y1-y2)
for any non zero complex number z=(x,y),there is a number z-1 such that zz-1=1, this multiplicative inverse
(x,y)(u,v)=(1,0) , so
xu-yv=1 ,yu+xv=0
Of linear simultaneous equations ,and simple computation yields the unique solution
u=x/(x2+y2) ,v=-y/(x2+y2) ,so the multiplicative inverse of z=(x,y) is
z-1=( x/(x2+y2) ,-y/(x2+y2)) (z?0)……………………(*)
the inverse z-1 is not defined when z=0 , in fact z=0 means that x2+y2=0
now ,to show(*)
z-1=1/z =1/(x+iy)=1/(x+iy) (x-iy)/(x-iy)=(x-iy)/(x^2+y^2 )=x/(x^2+y^2 )-iy/(x^2+y^2 )=(x/(x^2+y^2 ) , (-y)/(x^2+y^2 ))