Ima

Imaginary numbers : numbers that are built of real numbers that belong to the space R and an imaginary factor i multiplied by any given constant. It looks like this:

Imaginary numbers can be both treated and classified as two dimensional vectors – with a being the first coordinate and b being the second coordinate. A=(a,b) – is an imaginary number that has a real factor of a and an imaginary factor of b. You mean the real factor by writing: ReA=a And the imaginary factor by writing: ImA=b Its really easy and straightforward – if you have A=24 as a real number, its A=24 + i*0 as an imaginary number. The letter i in the expression is the anchor after which it is told how many imaginary digits you have to use. Some common imaginary numbers are: 1 3 WRITTEN AS AVECTOR : 1 , 3  i 2 2 2 2 1i WRITTEN AS A VECTOR :1 ,1 3  1 i WRITTEN AS AVECTOR : 3 , 1  2 2 2 2

Imaginary numbers undergo both algebra and vector equations. When you look at the chart above, you can see that if you draw a straight line from the point (0,0), you get a simple geometrical construct that can be viewed and described like this:

The fun starts: why not describe the number in another way? The length of the line, lets call it |Z|, defines the imaginary number. Why not use it? Bear in mind the Pythagorean theorem and a Pythagorean triangle. [Z ]= a 2b 2  a =cos ALFA [Z] b =sin ALFA [Z] Now, we can write down an imaginary number (1,1) as: [Z ]cosALFAisinALFA thus for1,1  isin   2cos  4 4 2 2      2cos 4 isin 4 = 2cos 4 2 isin 4 = 2  2  2i  2 =11i The notation above is used as commonly as (a,b) and a+ib. It helps to find higher powers and roots of an imaginary number. It's really important.

1 3 WRITTEN AS ∈THE GEOMETRICAL FORM :1cos  isin   i 2 2 3 3   1 3 1 cos isin = i 3 3 2 2 Straightforwad – cos pi/3 = ½ and sin pi/3=sqrt(3)/2 This leads us to the last form of an imaginary number, e to power: Z =e iArgZ 1 3 =  Arg  i 2 2 3 ArgZ is the angle that is used in the cosine and the sine. Thus:  Arg 1i= 4 Z =e iArgZ =[Z ] cos ArgZ isin  ArgZ =aib Where :[ Z ]= a2b2 ∧ ArgZ=angle How to count a hire power of an imaginary number? We use the geometrical form. In short, the module of the number is risen up the power we want to find and the argument of the number is multiplied times the power. Example: [ Z ]cos ArgZisin ArgZ 10=[ Z ]10 cos 10 ArgZ i sin 10 ArgZ  [ Z ]cos ArgZisin ArgZ 99=[ Z ]99 cos 99 ArgZ i sin 99 ArgZ  Finding the roots is analogous – you just have to remember that both sin and cos repeat themselves. So, a root of 3 in an imaginary number is: ArgZ n2  ArgZ n2  3 3 isin   Z  n=0= [Z ]cos  3 3 3 ArgZ n2  ArgZ n2  isin  3 Z  n=1= [Z ]cos  3 3 ArgZ n2  ArgZ n2  3 3 isin   Z  n=2= [Z ] cos 3 3 An imaginary numbers has as many roots as you want to find :). If you want to find 3 roots of an imaginary number, it will have 3 roots. If you want to find 4 roots, here's how it's going to look like: ArgZ n2  ArgZ n2  4 4 isin   Z  n=0= [Z ]cos  4 4 4 ArgZ n2  ArgZ n2  isin  4 Z  n=1= [Z ]cos  4 4 ArgZ n2  ArgZ n2  4 4 isin   Z  n=2= [Z ] cos 4 4 4 ArgZ n2  ArgZ n2  isin  4 Z  n=3= [Z ]cos  4 4

10 roots of 10 of imaginary number (1,0) - so of a real number 1 – looks like this: 10 10 [ Z ]=  120 2 =1 Arg 1,0=0 ArgZ n2  ArgZ n2  0 0 10 10 isin  =cos isin =1  Z  n=0= [ Z ] cos  10 10 10 10 10 ArgZ n2  ArgZ n2  2 2 10 isin  = cos isin =queer xD  Z  n=1= [ Z ]cos  10 10 10 10 ArgZ n2  ArgZ n2  4 4 10 10 isin  = cos isin =queer xD  Z  n=2= [ Z ]cos  10 10 10 10 10 ArgZ n2  ArgZ n2  6 6 10 isin =cos isin =queer xD  Z n=3 =  [Z ]cos  10 10 10 10 ArgZ n2  ArgZ n2  8 8 10 10 isin  =cos isin =queer xD  Z n= 4= [ Z ]cos  10 10 10 10 10 ArgZn2  ArgZ n2  10  10  10 isin  =cos isin =i  Z  n=5= [ Z ]cos 10 10 10 10 ArgZ n2  ArgZn2  12  12  10 10 isin =cos isin =queer xD  Z n=6 = [Z ] cos 10 10 10 10 10 ArgZ n2  ArgZ n2  14  14  10 isin =cos isin =queer xD  Z n=7 = [Z ]cos  10 10 10 10 ArgZ n2  ArgZ n2  16  16  10 10 isin =cos isin =queer xD  Z n=8 = [Z ]cos  10 10 10 10 10 ArgZn2  ArgZ n2  18  18 10 isin  =cos isin =queer xD  Z  n=9= [ Z ]cos  10 10 10 10



If you'd put in n=10, you'd recieve the same root as in n=0. So, that's it in a nutshell ;)