Transcript 11.3 Powers of Complex Numbers and De Moivre*s Theorem (de
11.3 POWERS OF COMPLEX NUMBERS AND DE MOIVRE’S THEOREM (DE MOI-YAY)
DeMoivre’s Theorem is used to raise complex numbers to integer powers.
If z= a+bi is any complex number with polar form rcis ϴ and n is any positive integer then the nth power of z is given by
z
n
n
r cis n
r
n
(cos
n
i
sin
n
)
The proof of this is simple, we remember that if (rcis ϴ) 2 And then by multiplying 2 complex numbers in polar form we get rrcis( ϴ+ ϴ) Which is the same as r 2 (cis2 ϴ) Now if we wanted to find (rcis Which is the same as r Which is the same as r 2 3 (cis2 cis3 ϴ)(rcis ϴ) ϴ.
= (rcis ϴ)(rcisϴ) ϴ) 3 . We would simply take (rcis ϴ) 2 (rcis ϴ).
Thus a pattern has developed that is consistent with de Moivre’s Theorem.
EXAMPLE
find
(2
cis o
15 ) 6
2
6
cis
o
64
cis
90
o
And in rectangular form that would be 0+64i Draw a picture to convince ourselves of this…
Rewrite this in rectangular form.
( 2
i
2) 4
Write in the form
a + bi
Let z=1-i Express z 3 , z 5 , z 7 in polar form. Then in rectangular form.
In order to complete this you must first put z into polar form.
z
2
cis
315
o
z
n
n
r cis n
z
So now we have 3 ( 2) 3
cis o
3(315 ) 2 2 2 2
cis cis
225 225
o o
In rectangular form that would be -8 – 8
i
On your own find the other 2
HWK PG. 410 1,2,3,5
11.4 ROOTS OF COMPLEX NUMBERS De Moivre’s Theorem helps as well to find roots.
The n nth roots of z =rcis ϴ are:
n z
z
1
n
1
n r cis
n
k
360
o n
for k
0,1, 2, 3, 4...
n
1
Add examples of taking roots of complex numbers in polar form.
We can also take nth roots of complex numbers, so maybe we want to find the square root or the cube root of (2+3i), DeMoivre’s Theorem is beneficial in helping us do that. Something to remember, if I want to find the fourth root of (2+3i), there will be in fact 4 different solutions that I could raise to the 4 th power to get (2+3i), so when you are asked for the nth roots of something you will have n solutions.
THE FOLLOWING COMES DIRECTLY FROM DE MOIVRE’S THEOREM
n z
z
1/
n
r
1/
n cis
n
k
2
n
,
for k
0,1, 2, 3...
n
1
Find the cube roots of 16i.
First understand that there will be a z 1 , z 2 , and z 3 .
To find z 1 you will use k=0 To find z 2 To find z 3 you will use k=1 you will use k=2 3
z
z
1/ 3 16 1/ 3
cis
3 2 3 16 1/ 3
cis
16 1/ 3 2 3
i
1 2 2.182247 1.25992
i