October 28, Complex Numbers.notebook. Discriminant

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1 OBJECTIVE Students will be able to utilize complex numbers to simplify roots of negative numbers. Students will be able to plot complex numbers on a complex coordinate plane. Students will be able to add and subtract complex numbers. Students will be able to multiply complex numbers. Students will be able to divide complex numbers using complex conjugates. Students will be able to find the absolute value of complex numbers. Discriminant Value greater than 0 and a perfect square greater than 0 and NOT a perfect square equal to 0 less than 0 Discriminant Type and Number of Roots rational roots irrational roots 1 rational root complex roots Complex roots result when the discriminant is negative. This will be our focus for today. Oct 4 10:58 AM Oct 4 11:06 AM Two Complex Roots Ax + Bx + C Discriminant is the expression b 4ac. The discriminant's value tells in advance how many answers and what type of answers to the quadratic formula there will be. x 5x + 7 = 0 ( 5) 4(1)(7) 5 4(1)(7) Two complex roots. Oct 4 11:37 AM Oct 4 11:07 AM You've been told that you can't take the square root of a negative number. 5 That's because you had no numbers which were negative after you'd squared them (so you couldn't "go backwards" by taking the square root). Every number was positive after you squared it. (+ 5) = + 5 ( 5) = + 5 So +5 is +5 and 5 or ±5 So you couldn't very well square root a negative and expect to come up with anything sensible. *Source Purplemath.com Now, however, you can take the square root of a negative number, but it involves using a new number to do it. This new number was invented (discovered?) around the time of the Reformation. At that time, nobody believed that any "real world" use would be found for this new number, other than easing the computations involved in solving certain equations, so the new number was viewed as being a pretend number invented for convenience sake. (But then, when you think about it, aren't all numbers inventions? It's not like numbers grow on trees! They live in our heads. We made them all up! Why not invent a new one, as long as it works okay with what we already have?) This new number was called "i", standing for "imaginary", because "everybody knew" that i wasn't "real". *Source Purplemath.com Oct 4 11:53 AM Oct 4 11:57 AM 1

2 i (imaginary number) = -1 i = ( -1 ) = - 1 i = -1 5 = 1 * 5 = i * 5 = ±5i 11 = 1 * 11 = i * 11 = ±11i So, let's see how this imaginary number is utilized. 48 = 1 * 48 = i * 16 * 3 = ±4i 3 Oct 4 1:01 PM Oct 4 1:15 PM Ax + Bx + C Discriminant is the expression b 4ac. The discriminant's value tells in advance how many answers and what type of answers to the quadratic formula there will be. x 5x + 7 = 0 ( 5) 4(1)(7) 5 4(1)(7) Two complex roots. Now let's solve for those roots. x 5x + 7 = 0 x = - (-5) ± (-5) - 4(1)(7) (1) x = - (-5) ± 5-8 (1) x = - (-5) ± - 3 (1) x = + 5 ± - 3 x = + 5 ± -1 * 3 x = + 5 ± i 3 Oct 4 1:35 PM Oct 4 1:36 PM 4x 3x + 3 = 0 Operations with Perform Operations with just as you would with a variable or monomial expression with a variable. Add imaginary numbers i + 3i. i + 3i = ( + 3)i = 5i Subtract imaginary numbers 16i 5i. 16i 5i = (16 5)i = 11i Multiply imaginary numbers (3i)(4i). (3i)(4i) = (3 4)(ii) = (1)(i ) = (1)( 1) = 1 Multiply imaginary numbers (i)(i)( 3i). (i)(i)( 3i) = ( 3)(i i i) = ( 6)(i i)=( 6)( 1 i) = ( 6)( i) = 6i Oct 4 1:46 PM Oct 4 1:38 PM

3 Pattern of Powers with i 1 i - 1 i 3 i i i 4 1 i 5 i 4 * i 1 i i 6 i 4 * i - 1 i 7 i 4 * i 3 i i 8 i 4 * i 4 1 Pattern of Powers with Simplify i 17 i 17 = i = i = i 1 = i Simplify i 10 i 10 = i 4 30 = i = i 0 = 1 Simplify i 64,00 i 64,00 = i 64,000 + = i 4 16,000 + = i = 1 Oct 4 1:45 PM Oct 4 1:58 PM Pattern of Powers with Simplify i 8 Simplify i 38 Simplify i 103 Simplify i 8,001 Complex Numbers A complex number is a number in the form of a + bi where a and b are real numbers. a is the real portion of the complex number. bi is the imaginary portion of the complex number. If b 0, then a + bi is an imaginary number i If a = 0 and b 0, then a + bi is a pure imaginary number i 6i Oct 4 1:15 PM Oct 4 1:30 PM Perform just as you would simplify like and unlike terms. Combine like terms. Leave unlike terms alone. Add Complex numbers (4 + i ) + (5 + 3i ) = 9 i + 3i = 5i 9 + 5i Subtract imaginary numbers (4 + i ) (5 + 3i ) (4 + i ) (5 + 3i ) (4 + i ) + ( 5 3i ) 4 5 = 1 i 3i = i 1 i Simplify (7 + 6i) + ( + 4i) Simplify (3 + i) + (9 + 8i) Simplify (1 5i) + (4 3i) Simplify (9 + 7i) (6 i) Simplify (5 i) (3 8i) Oct 4 1:06 PM Oct 4 1:18 PM 3

4 Multiply Complex numbers 9i (3 + i) 9i (3) = 7i 9i (i) = 18i 7i + 18( 1) 7i i Write in standard form. Multiply Complex numbers (4 + i ) (5 + 3i ) First (4)(5) = 0 Outer (4)(3i) = 1i Inner (i)(5) = 10i Last (i)(3i) = 6i 0 + 1i + 10i + 6i 0 + i + 6i 0 + i + 6( 1) Remember: i = i i (3 + i ) (6 + 8i ) (7 + 9i ) (5 + 4i ) Oct 4 1:3 PM Oct 4 3:01 PM Complex conjugates are to complex numbers that when multiplied together will result in a product that is a real number. (4 + 3i) (4 3i) First (4)(4) = 16 Outer (4)( 3i) = 1i Inner (4)(3i) = 1i Last (3i)( 3i) = 9i 16 1i + 1i 9i 16 9 i 16 9( 1) Remember: i = 1 Complex numbers must have real number denominators i i 3 i Allowed NOT Allowed Oct 4 :3 PM Oct 4 1:49 PM Dividing Complex numbers (4 + i ) (5 + 3i ) (4 + i ) * (5 3i ) (5 + 3i ) * (5 3i ) Multiply top and bottom by the complex conjugate of the denominator. First + Outer + Inner + Last (4)(5) + (4)( 3i) + (i )(5) + (i )( 3i ) (5)(5) + (5)( 3i ) + ( 3i )(5) + ( 3i )( 3i ) First + Outer + Inner + Last 0 i 6i 0 i 6( 1) 5 9i Remember: i = 1 5 9( 1) 0 i i i 17 ( + 3i ) (6 + 5i ) Oct 4 :4 PM Oct 4 :58 PM 4

5 (7 + 4i ) ( + i ) Oct 4 :59 PM Oct 4 3:06 PM (4 + 3i) 3i 4 (-5 + 4i) Oct 4 3:08 PM Oct 4 3:4 PM (- - 6i) The Absolute Value of a Complex Number is the distance the Complex Number is from the origin (0,0) on a coordinate plane. z = a + b where z is the complex number a + bi Oct 4 3:5 PM Oct 4 3:8 PM 5

6 Review DISTANCE FORMULA (x - x 1 ) + (y - y 1 ) Given A (, 5) B ( 7, 4), find AB. (-7-4) + ( - -5) (-11) + (7) (11) + (49) 170 Review DISTANCE FORMULA (x - x 1 ) + (y - y 1 ) Given C (4, ) D (3, 7), find CD. Oct 4 3:38 PM Oct 4 3:50 PM (x - x 1 ) + (y - y 1 ) complex number (5 + 3i) so (5, 3) origin (0, 0) (5-0) + (3-0) (5) + (3) z = a + b Remember Find the absolute value of the complex number ( + 7i). Oct 4 3:5 PM Oct 4 4:04 PM Find 4 + 9i. Oct 4 4:05 PM 6