Download Natasha deSousa MAE 501 Class Notes: 11/22 Up until today`s

Natasha deSousa
MAE 501
Class Notes: 11/22
Up until today’s lesson we said the domain of f (x)  x was all non-negative real numbers.
Now consider f : R  C , f (x)  x .
Will our function be defined on all the real numbers? Yes.


Next consider
 f : C  C , f (x)  x .


Can we find 2  2i ? Yes.
Algebraically:
(a  bi)(a bi)  2  2i


(a 2 b 2 )  2abi  2  2i
In order to understand how to complete this algebraic process, we must first understand
what a complex number is. A complex number is a number consisting of a real and an
imaginary part, written in the form x+yi, where x and y are real numbers. Complex numbers can
be defined as ordered pairs (x,y) of real numbers that are to be interpreted as points in the
complex plane, with rectangular coordinates x and y.1 The x is known as the real part and y is
known as the imaginary part. Therefore, in our equation above, we can set the real parts to be
equal and, likewise, set the imaginary parts to be equal. So, we now have:
2abi  2i , or ab 1
1
We have a system of equations. Take a  and plug this into our first
b
2
1 

b2  2
   
b 
1

 2  b2  2
b
1  b 4 2b 2
 2  2
b
b
4
 1  b  2b 2
equation.  b 4  2b 2 1  0
a2  b2  2

and
so,
2  4  4(1)(1)
2
2
 b  1  2
b2 
 b   1  2
1
Brown, James Ward and Ruel V. Churchill, Complex Variables and Applications (Boston:
McGraw-Hill, 2009) 1.

Since b must be a real number, we cannot have a negative number inside the radical; thus we are
left with b   1 2 . We have two choices for b, and, therefore, two choices for a.
We should get two square roots of 2  2i .
In the Reals:
x 2  2
x 2
x3  8

 x 3 8 2
we have 
two real solutions
we have one real solution
x 4  16

 x  4 16  2
we have two real solutions
x 5  243


we have one real solution
 x  5 243  3
We notice a pattern here: even powers yield two solutions and odd powers yield only one
solution.
QUESTION: Will this analogy follow through in the complex numbers?
(Some students said yes, others suggested there would be an infinite number of roots.)
Now let’s look at finding the square root of 2  2i by converting to polar coordinates.
The polar coordinate system is a two-dimensional coordinate system in which each point on the
plane is determined by the distance from a fixed point and an angle from a fixed direction. 2 To
convert between rectangular and polar
 coordinates, (x, y)  (r, ) , we first recognize, in the
graph below, we have a right triangle. We can obtain the equations: x  rcos and y  rsin  .

2
http://en.wikipedia.org/wiki/Polar_coordinate_system


We can use the following equations to easily convert from rectangular to polar coordinates:
thus,
thus,
Let (r, ) be the polar coordinates of the point (x, y) that corresponds to a non-zero complex
number z  x  yi . This complex number can be written in polar form as
y 
z  r(cos   isin  ) , where r  z  x 2  y 2 and   tan 1 
x 


(Note:  has an infinite number of possibilities, thus we choose      . The value of  is
also determined by the quadrant containing the point z.)



By Euler’s formula, we have e i  cos   isin  .



This equality can be proven by using the Taylor series expansions of e x ,cos x,and sin x (where x
is a real number). The expansions are as follows:


Using the knowledge of the powers of i, and the series above, we obtain the expansion of e ix

Thus, we can write z  r(cos   isin  ) in exponential form:
z  re i


Now let’s go back to our example and convert 2  2i to polar form. Notice that we can first
simplify this expression and obtain 2(1 i) .
So, x  1 and y  1
We have:



r  x 2  y 2
  tan 1
 r  12 12
1
1
1
   tan (1)
   tan 1
r 2
 

So,
y
x


4
4

2(1 i) = 2 2(cos  isin )  2 2e 4

4
i

SOME PROPERTIES:

i
i
When multiplying two complex numbers, z1  r1e and z 2  r2e , we obtain

z1z2  r1e i r2e i  r1r2e i e i  (r1r2 )e i(  ) .
*We add the two angles because:


e i e i  (cos   isin  )(cos   isin  )
 (cos  cos   sin  sin  )  i(sin  cos   cos sin  )
 cos(   )  isin(    )
 e i(  )
When raising a complex number to a power, n, we obtain:

z n  (re i )n  r ne in
Now to find the square roots of z  2 2i , using polar form:
1
2



i
1
2
1
2
 1 
z  (2 2e )  (2 2) (e
4
 i
4 2 

i

i
)  2 2e  8e 8  4 8 (cos
8
4


 isin )
8
8

Is this the only square root of z  2 2i ? No.
i
i
First observe that two non-zero complex numbers, z1  r1e and z 2  r2e are equal if
and only if r1  r2 and     2k , where k is some integer.3 An nth root of any non-zero
i
complex number, z 0 r0e , is a nonzero number

3


Brown24-25.
z  re i such that z n  z0 , or r n e in  r0e i




For these two complex numbers to be equal, by the statement above, the following must hold:
r n  r0 and n    2k , where k is any integer
then, r  n r0 and


c k  n r0 e
So,

i(

n


2k
)
n

 2k

n
n
are the distinct nth roots of z 0
(Note: All distinct roots are found when k = 0, 1, 2, …., (n-1); otherwise, we arrive at the same
points and begin repeating roots. )


Using this formula, we can now find the square roots of z  2 2i . We will only use k = 0 and
k = 1. Since we are finding the square root, n = 2. The distinct roots are:
c 0  2 2e
c1  2 2e
 1 2(0)
i( )( )
4 2
2
  1 2(1) 
i( )( )

 4 2
2 
 2 2e

 2 2e

i
8

i  
8 
 8e
4
 4 8e
i
i

8
9
8
 4 8 (cos

 isin

)
8
9
9
 4 8 (cos  isin )
8
8
8

(We stop after k = 1. If we used k = 2, we would have obtained, 4 8 (cos 18  isin 18 ) , which is
 point as
the same
4
8 (cos

8
8

8
 isin ) on the complex plane; they are equal as complex numbers.)
8


EXERCISES:
1. Express the following complex numbers in both polar and exponential form
a) (1  i)
b) (1 3i)

2. Find the fourth (complex) root of -9.

3. After learning how to find the nth root of a complex number, does the analogy of even
and odd powers determining the number of solutions in the reals apply to the complex
numbers? If not, is there a pattern and what is it?
ANSWERS:
1. a) x = 1 and y =-1 so, r  12  1  11  2 and   tan 1
2
thus (1  i)  2(cos


i


 isin
)  2e 4
4
4
b) x =1 and y = 3 so, r  1 
2

thus (1 3i)  2(cos

3

 3
2
 isin )  2e
3
i
1

 tan 1 (1) 
1
4

3

 tan 1 3 
 1 3  4  2 and   tan 1
1
3

3



2. First, express -9 as a complex number then convert it into exponential form:
9  9  0i  9(cos   isin  )  9e i

n = 4 and we will use k =0,1,2 and 3.
So, using our formula to find the nth roots, we have:
c 0  9e


1 2(0)
i( )( )
4
4

i

 9e 4  4 9 (cos


 isin )
4
4
1 2(1)
3
i( )( )
i
3

3
c1  4 9e 4 4  4 9e 4  4 9 (cos
 isin )
4
4
1 2(2)
5
i( )( )
i
5
5
c 2  4 9e 4 4  4 9e 4  4 9 (cos
 isin )
4
4
1 2(3)
7
i( )( )
i
7

7

c 3  4 9e 4 4  4 9e 4  4 9 (cos
 isin
)
4
4
4
4


3. No, the analogy for the complex numbers is not the same as the reals. The number of
solutions is equal to the power. The square root has two solutions, the cube root has three
solutions, the fourth root has four solutions, etc.