De Moivre Flashcards
How to prove de moivre theroem for negative integers?
Let n = - p where p positive integer
So it becomes 1 / etc ^p, where you can now use de moivre
Now RATIONALISE by multiplying by complex conjugate (which is neagtiveangle but by odd even just - isintheta )
Bottom simplifies to 1, top is calm
Now sub in n for p, so -n
Use odd even again to show it becomes the same
And thus works for all negative integers n
How to prove de moivre for positve by induction
Show true for n =1 is easy
Assume true for n = k cal
Prove true for n = k+1,
- multiply indices
- expend, collecte real and imaginary oarts (fsctor out I )
- two compound angle formuale here to use, and shows it’s true
Now say conclusion, as true for j= 1 and if true for n =k it’s true for n=k+1 thus true for all integrated greteatrr than n
Conclusion for negative de moivre
True fir negative interest assuming true for positive integ
Remember how to do wacky de moivre?
Either use odd even
Or DIVIDE BY I, to make it in the form cos + isin rather than sin + icos
And manipulate !
How to simplify a indict like root 8 to -7
As it shows decimal, do 1/ it first
Then do an even term to get an even number, and then add on the root
Double check
This is how to leave answers in EXACT FORM
Why sum if roots of unity = 0
Can be shown as 1 , W , W 2, w3 if W is deifned as the unity with lowest positive arg
Using n = 2piK/n m where k is 0 to n-1
Thus is shown
Geometric sequence adds up
W ^n = 1 as W represntsa root of number z where z ^n was 1 initially
Everything need to know about complex roots of UNITY and in general
Unity =1
In General is something random
Roots of unity can be found using same technique with mod 1, but wadier to just use data sheet method
- these will al, o,it to form a shape, and there will always start on x axis as one root is 1
- sum of roots sum to 0 as you can represent the smallest positive arg as W , and write all the roots in terms of W , 1, W, w2 etc
Now do gemotric addton and rememebr W ^n always is 1 as any root to the n is 1
For roots of a complex number
- also forms a shape
- easiest way is to use mod z e to the theta, and work out all the thetas from principle to highest
- will form shale but slightly moved, angles in between still = 2Pi/n!
EASISEST WAY TO SOLVE ANY ROOT OF COMPLEX NUMEBR PROBLEM
Either find all the arguments, but this could be long
Or write the equation down once and use big calc to do it
So write down exactly what it will be, and remember to now add 2Pi (x) / n too each time
In calc jus change this each time and see what you get!
If they say sqaure centred at 3 -5i with one corner at origin how to find equation
Each corner is always a solution. So if they tell you the position of the solution, subbing in and edoanidmg will give you RHS
If centred at 3 -5i must do -( 3-5i)
As sqaure muktiply to power if 4!
When dividing in arg cos form, how does the rules still apply?
Still divide mucus and subtrsvt arg, same works but can see it more clear,eh using de movire theorem
