Download Lesson One Complex Numbers File

Complex Numbers
i  1
2
i  i
3
i  1
4
Simplify:
Answer:
(2  3i)  (4  5i)
(2  3i )
 6  8i
 5  12i
2
i i i
i
(7  2i)(7  2i)
 53
12
13
14
Simplify:
2i 5
3i 5
Answer:
1
14
Simplify:
i  2i
i
2
3
Answer:
2  i

5 5
14
i
If z = 1 – 2i, simplify the following:
a)
z 1
2
z
b)
z2
z 3i
Answer:
2
25
 14
25 i
Answer:
18
10
 101 i
We usually use the letter z to represent a complex number.
z  x  yi
The Real portion of z is given as Re(z) = x.
The Imaginary portion of z is given as Im(z) = y.
If
z  x  yi
what is
z* or z
If
z  x  yi
z
z* or z ?
is the complex conjugate of z.
what is the modulus of z, or
is the modulus and is always positive or
z?
z  x y
2
2
Graphing Complex Numbers
use the Argand Diagram or the Complex Plane.
y-axis would be the imaginary numbers
Im
5
4
3
2
1
5
4
3
2
1
1
1
2
3
4
5
2
3
4
5
Re
x-axis would be the
real numbers
Graph the complex number z = 2 + 3i
Graph the complex number z = -4i
Im
5
4
3
2
1
5
4
3
2
1
1
1
2
3
4
5
2
3
4
5
Re
4
Im
P(x, y)
3
2
r
y
1

-4
-2
2
x
4
6
-1
If z = x + yi
-2
2
2
z

x

y
then modulus is:
-3
and the argument is: arg( z )  
-4
y
where tan  
x
Re
P(r cos  , r sin  )
4
Im
P(x, y)
3
2
r
y
1

-4
-2
2
x
4
If z = x + yi
-1
notice also that
x  r cos
and
y  r sin 
-2
therefore
z  x  yi  r cos  ir sin   r (cos   i sin  )
-3
-4
this is called the modulus-argument form of a complex number.
6
Re
8
Notice also that:
z  r (cos   i sin  )  rcis  rei
What if r = 1 and
   radians?
cos   i sin   ei
ei  1
i
e 1  0
This is known as Euler’s Formula and was
discovered in 1748. It is considered by many to
be the most beautiful of all formulas since it
combined so many different numbers together into
one simple formula.
Write the following in modulus-argument form:
z1  2  2 3i
 
z1  4  cis 
3

z2  1  i
 3 
z2  2  cis

4 

z3  3
z3  3  cos  i sin  
Write the following in modulus-argument form:
z1  3 3  3i



z1  6  cos  isin 
6
6

z2  2  2i
3
3 

z2  2 2  cos
 i sin

4
4


z3  4i

 

z3  4  cos
 i sin

2
2 

If
z1  r1 (cos1  i sin 1 )
what is
and
z2  r2 (cos 2  i sin 2 )
z1  z2 ?
z1  z2  r1  r2 [cos(1  2 )  i sin(1  2 )]
To find the product of two complex numbers, you multiply their
moduli and add their arguments.
If
z1  r1 (cos1  i sin 1 )
what is
and
z2  r2 (cos 2  i sin 2 )
z1
?
z2
z1 r1
 [cos(1   2 )  i sin(1   2 )]
z2 r2
To find the quotient of two complex numbers, you divide their moduli
and subtract their arguments.
Given the following complex numbers below, find the following in
modulus-argument form:
z  1 i
w  1  i 3
a)
zw
11
z  w  2 2cis
12
b)
z
w
z
2
5

cis
w
2
12
Given the following complex numbers below, find the following in
modulus-argument form:
z  3  3i
w   3i
a)
zw
7
z  w  6 2cis
12
b)
w
z
w
2 13

cis
z
3
12
Given the following complex numbers below, find the following in
modulus-argument form:
z  sin
a)

6
 i cos
zw
z  w  cis

12

6
w  cos
b)

4
 i sin
w
z
w
7
 cis
z
12

4
If:
z1  z2  r1  r2 [cos(1  2 )  i sin(1  2 )]
then, what is?
z  r12 [cos(21 )  i sin(21 )]
2
1
z  r13[cos(31 )  i sin(31 )]
3
1
z  r1n [cos(n  1 )  i sin(n  1 )]
n
1
This is known as DeMoivre’s Theorem
z n  [r (cos   i sin  )]n  r n (cos n  i sin n )
 r cis n  r n ein
n
If z = 1 + i, find
z4
Answer:
Find the value of
Answer:
4cis  4
1
(4  4i )3
2
cis  34 
256
Find the value of
  
  
cos  6   i sin  6  
 
  
Answer:
1
3
  3
 cos 
  4

 3
  i sin 

 4



2
If z = x + yi, find x and y if:
z  2i  2 z  2 z  3  i
*
x = -3, y = -1
Find the modulus and argument of:
(cos 4  i sin 4 )
z

 3
[2(cos 3  i sin 3 )]

mod  18 , arg  2
 2
Given that
z  (b  i)2 , where b is real and positive, find the exact
value of b when
Answer:
Let
arg z  60 .
b 3
z1 and z2
be complex numbers. Solve the simultaneous equations:
2 z1  3z2  7, z1  iz2  4  4i
Give your answers in the form
Answer:
z1  2  3i
z2  1  2i
z  a  bi
where
a, b  Z .
Given:
 cos   i sin  
6
a. Expand using the Binomial Theorem.
b. Expand using de Moivre’s Theorem.
c. By equating the real and imaginary parts, show that:
sin 6
 a cos5   b cos3   c cos 
sin 
and find the values of a, b, and c.
Answer:
a = 32, b = -32 and c = 6