11.3 Powers of Complex Numbers, DeMoivre's Theorem Objective To use De Moivre’s theorem to find...
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Transcript of 11.3 Powers of Complex Numbers, DeMoivre's Theorem Objective To use De Moivre’s theorem to find...
11.3 Powers of Complex Numbers,DeMoivre's Theorem
Objective To use De Moivre’s theorem to find powers of complex numbers.
(r cisθ)³ = (r cisθ)(r cisθ)(r cisθ)
Remember: multiply absolute values and add polar angles.
= (r² cis 2θ) (r cisθ)
= r³ cis 3θ
(r cis θ)ⁿ = rⁿ cis nθ
Abraham de Moivre(1667 - 1754)
DeMoivre’s Theorem
If is a complex number,
then
z r i cos sin
z r n i nn n cos sin integer. positive a is 1 where n
You can repeat this process raising complex numbers to powers. Abraham DeMoivre did this and proved the following theorem:
z r n i nn n cos sin
This says to raise a complex number to a power, raise the modulus to that power and multiply the argument by that power.
6
54sin
6
54cos24
i
43 i
r 3 1 4 22 2
4
4
6
5sin
6
5cos23
ii
This theorem is used to raise complex numbers to powers. It would be a lot of work to find
iiii 3333 you would need to FOIL and multiply all of these together and simplify powers of i --- UGH!Instead let's convert to polar form
and use DeMoivre's Theorem.
3
1tan 1 but in Quad II
6
5
3
10sin
3
10cos16
i
i
2
3
2
116
8 8 3i
Solve the following over the set of complex numbers:
13 z We know that if we cube root both sides we could get 1 but we know that there are 3 roots. So we want the complex cube roots of 1.
Using DeMoivre's Theorem with the power being a rational exponent (and therefore meaning a root), we can develop a method for finding complex roots. This leads to the following formula:
n
k
ni
n
k
nrz n
k
2sin
2cos
1 , ,2 ,1 ,0 where nk
101 22 r
n
k
ni
n
k
nrz n
k
2sin
2cos
Let's try this on our problem. We want the cube roots of 1.
We want cube root so our n = 3. Can you convert 1 to polar form? (hint: 1 = 1 + 0i)
01
0tan 1
We want cube root so use 3 numbers here
Once we build the formula, we use it first with k = 0 and get one root, then with k = 1 to get the second root and finally with k = 2 for last root.
2,1,0for ,3
2
3
0sin
3
2
3
0cos13
k
ki
kzk
10sin0cos1 iHere's the root we already knew.
3
12
3
0sin
3
12
3
0cos13
1
iz
ii2
3
2
1
3
2sin
3
2cos1
3
22
3
0sin
3
22
3
0cos13
2
iz
ii2
3
2
1
3
4sin
3
4cos1
If you cube any of these numbers
you get 1. (Try it and see!)
3
02
3
0sin
3
02
3
0cos13
0
iz
2,1,0for ,3
2
3
0sin
3
2
3
0cos13
k
ki
kzk
ii2
3
2
1,
2
3
2
1,1 We found the cube roots of 1 were:
Let's plot these on the complex plane about 0.9
Notice each of the complex roots has the
same magnitude (1). Also the
three points are evenly spaced
on a circle. This will always be
true of complex roots.
each line is 1/2 unit
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.
Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au