Complex numbers De moivres theorem KUS objectives BAT
Complex numbers: De moivre’s theorem KUS objectives BAT know how prove and apply De. Moivres theorem Starter:
De Moivre Theorem’
BASIS Proof (1) De Moivre’s theorem can be proved using the method of proof by induction from FP 1 Proving that: is true for all positive integers ASSUMPTION Basis – show the statement is true for n = 1 INDUCTIVE Assumption – assume the statement is true for n = k Inductive – show that if true for n = k, then the statement is also true for n = k + 1 Conclusion – because the statement is true for n = 1 and also true if any value is, then the statement is true for all values of n You can write this as two separate parts as the powers are added together CONCLUSION We can rewrite the first part based on the assumption step, and the second based on the basis step Using the multiplication rules from 3 B Multiply the moduli, add the arguments As all the ‘k’ terms have become ‘k + 1’ terms, if the statement is true for one term, it must be true for the next, and so on… The statement was true for 1, so must be true for 2, and therefore 3, and so on… We have therefore proven the statement for all positive integers!
Proof (2) (cont) We have just proved theorem for n = k where k is a positive integer Now we need to show it is also true for any negative integer If n is a negative integer, it can be written as ‘-m’, where m is a positive integer Write using a positive power instead Use De Moivre’s theorem for a positive number (which we have proved) Multiply to change some terms in the fraction Multiply out like quadratics – the bottom is the difference of two squares i 2 = -1 You can see that the answer has followed the same pattern as De Moivre’s theorem! You cancel the denominator as it is equal to 1 Use cos(-θ) = cos(θ) and sin(-θ) = -sinθ
Proof (3) (cont) Having now proved that De Moivre’s theorem works for both positive and negative integers, there is only one left We need to prove it is true for 0! This is straightforward. As it is just a single value, we can substitute it in to see what happens Sub in n = 0 Left side = 1 as anything to the power 0 is 1 You can find cos 0 and sin 0 as well ‘Calculate’ So we have shown that De Moivre’s Theorem is true for all positive integers, all negative integers and 0’ It is therefore true for all integers!
Theorem It is important to note that De Moivre’s theorem can also be used in exponential form. Both parts will be raised to the power ‘n’ You can remove the bracket! This is De Moivre’s theorem in exponential form!
The denominator has to have the ‘+’ sign in the middle Apply cos(-θ) = cosθ and sin(-θ) = -sinθ Apply De Moivre’s theorem (there is no modulus value to worry about here!) Just multiply the arguments by the power Apply the rules from 3 B for the division of complex numbers Divide the moduli and subtract the arguments Simplify the sin and cos terms Calculate the sin and cos terms Simplify
y You need to write this in one of the forms above, and you can then use De Moivre’s theorem r √ 3 Start with an argand diagram to help find the modulus and argument of the part in the bracket θ 1 x
KUS objectives BAT know how prove and apply De. Moivres theorem self-assess One thing learned is – One thing to improve is –
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