Download VII) ARITHMETIC WITH COMPLEX NUMBER

College Algebra to Calculus and the TI-83
Lesson 3
Complex Numbers
I. ARITHMETIC WITH COMPLEX NUMBER
To enter the complex number a + bi, use the i key ( 2nd
)
Exercise 1. Perform the indicated operations with complex numbers
(3 - 2i) + (-3 +5i) - (0 - 3i) - (-5 + 4i)

(3 - 2i) + (-3 +5i) - (0 - 3i) - (-5 + 4i)
ENTER
answer: 5 + 2i
Exercise 2. Multiply (2+3i)(3-5i)



ON
(2+3i)(3-5i)
ENTER
answer: 21- i
6
Exercise 3: find (3  4i)
 (3+4i)^6
 ENTER
answer: 11753 – 10296i
Exercise 4: Divide 5 by 1+2i


5(1+2i)
ENTER
answer: 1 -2i
Exercise 5: Compute (5-12i)  (-1 -2i)


(5 - 12i)  (-1 – 2i) ENTER
MATH 1 ENTER
4  2i
1 i
(2 + 3i)(5 - 2i) + (2 - i)^3 + (4 + 2i)  (1 - i) ENTER
Exercise 6.

ans.: 3.8 + 4.4i
answer: 19/5+22/5 i
Simplify
(2  3i )(5  2i )  (2  i )3 
answer: 19 + 3i
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1
2
Exercise 7. Simplify a) i b)i c) i d ) i e) i 11 f ) i  g ) e i
 a) i^2
ENTER
anwer.: -1
 b) i^5
ENTER
answer: i
 c) i^-1
ENTER
answer: -i
 d) i^(12) ENTER
answer: 0 .707 + 0.707 i (rounded to 3 decimals)
 e) i^-11 ENTER
answer: 1E-13 + i or i
 i^π
ENTER
answer: 0 . 221 - 0.975 i
 e^(i  )
ENTER
answer: -1
2
1
5
II. Using the Complex MODE
MODE        a+bi ENTER
Exercise 8. Express in the form a+bi
a)  9
b) (2   4 )(1  3  16 )

MODE       

2nd

(2 + 2nd
(- 9 )
ENTER
a + bi ENTER 2nd QUIT
answer: 3i
(-4)) (1 - 3 2nd
answer: 26 - 22i
(-16))
3
2
Exercise 9. Evaluate the expression x  2x  19x  35 for x = 2 - i
 (2 - i)^3 + 2(2 - i)^2 -19(2 - i) +35 ENTER
answer: 5
Another method:
 2 – i STO x
 x^3 + 2x^2 - 19x + 35 ENTER Answer: 5
Exercise 10. Find all solutions of the equation x 2  4 x  13  0
 MODE       
a + bi ENTER 2nd QUIT
 1 STO ALPHA A : -4 STO ALPHA B : 13 STO ALPHA C
 ENTER

(-B +
ENTER
answer: 2 + 3i

2nd ENTRY (change the + to -) ENTER
answer: 2 – 3i
(B^2 - 4AC))(2A)
II. Representation of complex numbers as vectors in the complex plane (optional)
Consider the complex number a + bi
The complex numbers menu: MATH  CPX
1: conj( - gives the conjugate a – bi of a complex number a + bi
Exercise 11. Find the conjugate of 5 + 3i
 MATH  CPX ENTER conj(5+3i) ENTER answer: 5 – 3i
Exercise 12. Find the conjugate of (3 – 2i)(5+7i)
 MATH  CPX ENTER conj((3 – 2i)(5 + 7i)) ENTER
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answer: 29 – 11i
2: real( -gives the real component “a “ of a complex number a + bi
Exercise 13. Find the real component of the complex number (5  7i ) 5

MATH  CPX
2 ENTER real(5 – 7i)^5
ENTER
answer: 1900
(2  i) 2
Exercise 14. Find the real component of the complex number
5  3i

MATH  CPX
answer: 3/34
2 ENTER real((2 + i)^2  (5-3i) ) MATH 1
ENTER
3. imag( gives the imaginary component b of the complex number a + bi
Exercise 15. Find the imaginary component of the complex number

MATH  CPX
answer: 29/34
(2  i) 2
5  3i
3 ENTER imag((2 + i)^2  (5-3i) ) MATH 1
ENTER
4. angle( -given the angle that the segment from the point (0, 0) to the point (a, b) makes
with the positive end of the x-axis. The angle is also called the argument  of the
complex number.
Exercise 16. Find the argument of the complex numbers a) 3 + 4i b) -3 + 3i
3 -i (set the calculator in degree mode)
c) -1- 3 i
d)
 MODE   Degree ENTER 2nd QUIT



MATH  CPX 4 angle(3 + 4i) ENTER answer: 53.13 degrees.
2nd ENTRY (change 3 + 4i to -3 + 3i) ENTER answer: 135 degrees
2nd ENTRY (change -3 + 3i to -1- (3 )i ) ENTER answer: -120 degrees

2nd ENTRY (change -3 + 3i to
(3 ) -i ) ENTER answer: -30 degrees
5. abs (-gives the magnitude or absolute value or modulus r of the complex number, that
is, the distance from the point (0, 0) to the point (a, b)
Exercise 17. Find the magnitude or modulus of the complex numbers a) 3 – 4i, b) 2+5i
c) -5 – 12i
 MATH  CPX 5 abs(3 – 4i) ENTER answer: 5.
 2nd ENTRY (change 3 – 4i to 2 + 5i) ENTER answer: 5.385164807
 2nd ENTRY (change 2 + 5i to -5 – 12i) ENTER answer: 13
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Note: the form a + bi is called the rectangular form of a complex number. The form
rei is called the polar form of the complex number where r is the modulus or absolute
value of the complex number and  is the angle or argument.
6. ►Rect -converts a complex number to rectangular form, that is, given r and 
obtain the complex number in the form a + bi.  is always in radians. When  is given
in degrees, multiply  by

180
to convert to radians.
Exercise 18. Find the complex number with modulus 20 and argument 150 degrees.
 20e^(150  i  180)
 MATH  CPX 6 ►Rect ENTER answer: -17.3205+10i
or
 20e^(150  i  180) ENTER answer: -17.3205+10i
Exercise 19. Find the complex number with modulus 5 2 and argument 225 degrees.

5 (2) e^(225  i  180)
 MATH  CPX
or

6 ►Rect ENTER
5 (2) e^(225  i  180) ENTER
answer: -5-5i
answer: -5-5i
7. ►Polar - converts a complex number a + bi to the polar form re^(i) where
 is in radians if the calculator is in radian MODE, it is in degrees if the calculator is in
degrees MODE
Exercise 20. Convert to polar form in degree mode: a) -2+2i
 MODE   Degrees ENTER 2nd QUIT
 -2+2i MATH  CPX 7 ►Polar ENTER
b) 3 – 4i
c) -1+
3i
answer: 2.828427125e^(135i) or r=2.828 =2 2 and =135 degrees
 2nd ENTRY (change -2 + 2i to 3 – 4i) ENTER
answer: 5e^(-53.13 i) or r=5 and  = -53.13 degrees
 2nd ENTRY (change 3 – 4i to -1+ 3 i ) ENTER
answer: 2e^(120 i) or r=2 and =120 degrees
Exercise 21. Convert the complex number
2  3i
to polar form, find the modulus and
1  2i
the argument in degrees.
 MODE   Degrees ENTER 2nd QUIT
 (2 – 3i)  (1  2i ) MATH  CPX 7 ►Polar ENTER
answer: 1.61245155e^(-119.7448813i) or r= 1.61 and =-120 degrees approximately-21-
III. Using complex numbers to create Pythagorean triplets (optional)
Definition: a Pythagorean triplet is a set of natural number {a, b, c} such that
a 2  b 2  c2 .
How to create Pythagorean triplets?
One way to create Pythagorean triplets is:
Step 1. Let x and y be any natural numbers, x>y
Step 2. Consider the complex number x + y i
2
2
2
Step 3. Compute (x  yi)  a  bi and c  a  b =abs(a + bi)
answer: {a, b, c} is a Pythagorean triplet.
x and y are called the generators of the Pythagorean triplet {a, b, c}
Exercise 21. Create the Pythagorean triplet generated by the numbers 8 and 5, using the
TI-84
 (8 + 5i)^2 ENTER
answer: 39 + 80i
 MATH  CPX 5
 abs( 2nd ANS )
ENTER
answer: 89
answer: the numbers 39, 80, 89 form a Pythagorean triplet, that is 39 2  80 2  89 2
Exercise 22. Create a Pythagorean triplet using the generators 23 and 17



(23 + 17i)^2 ENTER
answer: 240 + 782i
MATH  CPX 5
abs(2nd ANS)
ENTER
answer: 818
answer: the numbers 240, 782, 818 form a Pythagorean triplet
Exercise 13. Create a Pythagorean triplet using the generators 4 and 3.
 (4 + 3i)^2 ENTER
answer: 7 + 24i
 MATH  CPX 5
 abs( 2nd ANS )
ENTER
ans.: 25
answer: the numbers 7, 24, 25 form a Pythagorean triplet
Exercise 14. Create a Pythagorean triplet using the generators 25 and 24.
 (25 + 24i)^2 ENTER
answer: 49 + 1200i
 MATH  CPX 5
 abs( 2nd ANS )
ENTER
answer: 1201
answer: the numbers 49, 1200, 1201 form a Pythagorean triplet
Note: in exercises 13 and 14 it can be observed that if the generators x and y are
consecutive natural numbers then b and c are also consecutive.
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Theorem. If x and y are any two natural number, x>y,
a  bi  ( x  yi ) 2 and c  a 2  b 2 , then c  x 2  y 2 and the set {a, b, c}
forms a pythagorean triplet
Proof of theorem:
1. ( x  yi ) 2  ( x  yi )( x  yi )  x 2  xyi  xyi  y 2 i 2  x 2  y 2  2xyi  a  bi
 a  x 2  y 2  0, b  2xy  0 where a and b are natural numbers .
2. Let c  a 2  b 2
3.
 c 2  a 2  b 2  ( x 2  y 2 ) 2  ( 2xy) 2  x 4  2x 2 y 2  y 4  4x 2 y 2
 ( x 2  y 2 ) 2  c  x 2  y 2 , c  0, where c is a natural number .
Therefore a, b and c are natural numbers such that a  b  c where c is the modulus
2
of the complex number ( x  yi )
2
2
2
Theorem. If x and y are two consecutive natural number, y= x-1 and they generate the
Pythagorean triplet {a, b, c}, then b and c are consecutive , c=b+1.
Proof:
1. x and y=x-1 are generators of the Pythagorean triplet {a, b, c} 
2
2
2
2. b= 2xy = 2x(x-1) = 2 x  2 x and c  x  y by previous theorem and
therefore
2
c  x 2  y 2  x 2  x  1  2 x 2  2 x  1 = b+1
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