Alg1131W6T4 - Complex conjugate root theorem
Saturday, 14 April 2018
12:15 PM
Created with Microsoft OneNote 2016.
![Untitled picture.png Key properties of complex conjugation that we discussed in the last topic were the rules
3) (zw)
and
and
w)
There is one more rule that will be useful to state explicitly:
This means that a complex number z is actually a real number precisely when z
to demonstrate step by step. Suppose that z is a solution to the equation
Then if we take the complex conjugate of both sides, we get
The justification for this step is [select all that apply]:
Rule 1
So then
which is justified by
Rule 1
Next we rewrite this as
Rule 2
Rule 2
which is justified by [select all that apply]
Rule 1
So we get that
which is justified by
Rule 1
Rule 2
Rule 2
Rule 3
Rule 3
Rule 3
Rule 3
Rule 4
Rule 4
Rule 4
Rule 4
Rule 5
Rule 5
Rule 5
Rule 5
S. These rules have important consequences for solutions of polynomial equations which we are going
z +3z+8=0.
—o
+ 32 + 8
+ 32 8
Untitled picture.png Our conclusion is that if z is a solution to the equation; then so is S.
The reason this has worked so smoothly is that [select all that apply]
We know our rules for complex conjugation
this equation is a quadratic
this equation has small coefficients
this equation has real coefficients
we were lucky](Alg1131W6T4%20-%20Complex%20conjugate%20root%20theorem_files/image001.png)
![Untitled picture.png Our conclusion is that if z is a solution to the equation; then so is S.
The reason this has worked so smoothly is that [select all that apply]
We know our rules for complex conjugation
this equation is a quadratic
this equation has small coefficients
this equation has real coefficients
we were lucky
Untitled picture.png Recall that if the monic quadratic equation z + sz+ r = O has roots (or zeroes) WI and w2, then we can write
By comparing the coefficients of z on both sides, it follows that
WI tL'2
— 12
so:
2
i) The sum of the roots of z
0 is 4
O while the product of the roots is
-12
z
ii) The monic quadratic equation, in z, with complex conjugate roots 5 + 6i and 5 — 6i must be 0
iii) The monic quadratic equation, in z, with complex conjugate roots —5 + 2i and —5 — 2i must be O
- laz + 61
+ laz + 29
Untitled picture.png Machine generated alternative text:
i) If a monic quadratic polynomial in z has real coefficients, and one root is —4 + i, then that polynomial is
ii) If a monic quadratic polynomial in z has real coefficients; and one root is 1 — 4i, then that polynomial is
Untitled picture.png Machine generated alternative text:
Let's review how to transform the general quadratic polynomial
ax2 + bx + c.
(2
First we factor out a
Now add and subtract
0 to get
inside the brackets to get
2
b
use the syntax p/ (3 * q)
Note: to enter a fraction like
There is a perfect square inside these brackets; namely the square of
x+ö/(2a)
2
a
a
For the last step, we multiply through by
0 to get
a
a
2a) ¯ 4a2
4a
b2 _4ac
4a
This is one of the key derivations in mathematics, going back to the Persian mathematician al Khwarizmi (780-800 AD) We have rewritten the quadratic polynomial az2 + bz+c in the form
a(x — + k where
h
In fact the point [h, k] is the vertex ofthe parabola y = az2 + bz+ c.
-b/(2a)
- 4acV(4a)
and
+ k where
and
As a consequence, the parabola with equation y = 3m + 10m + 7 can be rewritten asy = 3 @
h
-10/6
This tells us that the vertex occurs at the point [-10/6,-4/31
Note: to write the point [7, 4] in Maple we write [7 , 4]
Ink Drawings
Ink Drawings

Ink Drawings
Ink Drawings
Ink Drawings
Ink Drawings
Ink Drawings
Ink Drawings](Alg1131W6T4%20-%20Complex%20conjugate%20root%20theorem_files/image002.png)
![Untitled picture.png Machine generated alternative text:
Let's review how to transform the general quadratic polynomial
ax2 + bx + c.
(2
First we factor out a
Now add and subtract
0 to get
inside the brackets to get
2
b
use the syntax p/ (3 * q)
Note: to enter a fraction like
There is a perfect square inside these brackets; namely the square of
x+ö/(2a)
2
a
a
For the last step, we multiply through by
0 to get
a
a
2a) ¯ 4a2
4a
b2 _4ac
4a
This is one of the key derivations in mathematics, going back to the Persian mathematician al Khwarizmi (780-800 AD) We have rewritten the quadratic polynomial az2 + bz+c in the form
a(x — + k where
h
In fact the point [h, k] is the vertex ofthe parabola y = az2 + bz+ c.
-b/(2a)
- 4acV(4a)
and
+ k where
and
As a consequence, the parabola with equation y = 3m + 10m + 7 can be rewritten asy = 3 @
h
-10/6
This tells us that the vertex occurs at the point [-10/6,-4/31
Note: to write the point [7, 4] in Maple we write [7 , 4]
Untitled picture.png Machine generated alternative text:
The usual quadratic formula follows from al Khwarizmi's identity: if
(b2 4ac)
az2 + c = z +
—o.
2a
then
4a
2 = b2 — 4ac, then we can solve for z in terms of a, b,c andr as
So if we can find a number r satisfying r
b
2a 2a
Hopefully you can recognize the usual quadratic formula here. If b2 — 4ac is a positive real number; we usually just replace r with •v/'j2¯7c.
However if b2 — 4ac is a negative real number, or in fact a general complex number—-which will happen if a, b and c are complex numbers—-then there is no canonical but
2 = b2 — 4ac, and then the formula z
we could still find a value r (at least approximately) satisfying r
Use this analysis to solve the complex quadratic equation
The solution set is
Note:
• make sure to simplify complex fractions
• the set {1 + i,2 — 3i} in Maple is written as {1 +1, 3
_b4_r —b—r
b
± — gives us the solutions (as a set):
2a 2a](Alg1131W6T4%20-%20Complex%20conjugate%20root%20theorem_files/image003.png)
