### 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 rcisϴ and n
is any positive integer then the nth power of z is given by
z  r cis n  r (cos n  i sin n )
n
n
n
The proof of this is simple, we remember that if (rcisϴ)2 = (rcisϴ)(rcisϴ)
And then by multiplying 2 complex numbers in polar form we get rrcis(ϴ+ ϴ)
Which is the same as r2(cis2ϴ)
Now if we wanted to find (rcis ϴ)3. We would simply take (rcis ϴ)2(rcis ϴ).
Which is the same as r2(cis2 ϴ)(rcis ϴ)
Which is the same as r3cis3 ϴ.
Thus a pattern has developed that is consistent with de Moivre’s Theorem.
EXAMPLE
o
find (2 cis15 )
2 cis (6  15 )
6
64 cis 90
o
o
And in rectangular form that would be 0+64i
Draw a picture to convince ourselves of this…
6
 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  r cis n
n
n
In rectangular form that would be
-8 – 8i
 So now we have
z  ( 2 ) cis 3(315 )
3
3
o
 2 2 cis 225
o
 2 2 cis 225
o
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 = rcisϴ are:
n
z  z
1 n
o

 k  360 
1 n
 r cis  
 for k  0,1, 2, 3, 4...n  1
n
n

 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 dif ferent 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
z  z
n
1/ n
r
1/ n
k  2 

cis  
 , for k  0,1, 2, 3...n  1
n 
n
Find the cube roots of 16i.
First understand that there will be a z1, z2, and z3.
To find z1 you will use k=0
To find z2 you will use k=1
To find z3 you will use k=2
3
z  z
1/ 3
 16
1/ 3
  2 0  2 
cis 


3 
 3
 16
1/ 3
 
cis  
 6 
 16
1/ 3
 3
1
i 

2 
 2
 2.182247  1.25992i
```