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# Sulalgtrig7e Isg 1 3

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### Sulalgtrig7e Isg 1 3

1. 1. a + bi
2. 2. When we take the square root of both sides of an equation or use the quadratic formula, sometimes we get a negative under the square root. Because of this, we'll introduce the set of complex numbers. i = −1 2 This is called the imaginary unit and its square is -1. We write complex numbers in standard form and they look like: a + bi This is called the real part This is called the imaginary part
3. 3. We can add, subtract, multiply or divide complex numbers. After performing these operations if we’ve simplified everything correctly we should always again get a complex number (although the real or imaginary parts may be zero). Below is an example of each. Combine real parts and ADDING (3 – 2i) + (5 – 4i) = 8 – 6i combine imaginary parts Be sure to distribute the SUBTRACTING (3 – 2i) - (5 – 4i) negative through before combining real parts and 3 – 2i - 5 + 4i = -2 +2i imaginary parts FOIL and then combine like MULTIPLYING (3 – 2i) (5 – 4i) terms. Remember i 2 = -1 = 15 – 12i – 10i+8i2 Notice when I’m done simplifying that I only have two terms, a real =15 – 22i +8(-1) = 7 – 22i term and an imaginary one. If I have more than that, I need to simplify more.
4. 4. 3 − 2i 5 + 4i = + 12i − − i − ( 8i ) 1515 + 12i1010i8−− 12 DIVIDING ⋅ = 5 − 4i 5 + 4i + + i − − i − −( −i ) 25252020i2020i161612 FOIL Combine like terms i 2 = −1 To divide complex numbers, you multiply the top and bottom of the fraction by the conjugate of the bottom. 23 + 2i 23 2 This means the same = = + i complex number, but 41 41 41 with opposite sign on the imaginary term We’ll put the 41 under each term so we can see the real part and the imaginary part
5. 5. Let’s solve a couple of equations that have complex solutions. Square root and x + 25 = 0 2 x = ± − 25 2 don’t forget the ± -25 -25 x = ± 25( − 1) = ± 25 − 1 = ±5 i The negative 1 under the square root becomes i Use the x − 6 x + 13 = 0 2 − b ± b 2 − 4ac quadratic formula x= 2a − ( − 6) ± ( − 6) − 4(1)(13) 2 6 ± 36 − 52 x= = 2(1) 2 6 ± − 16 6 ± 16 i 6±4i = = = = 3 ± 2i 2 2 2
6. 6. Powers of i We could continue but notice i=i that they repeat every group i = −1 2 of 4. For every i 4 it will = 1 i = i i = −1(i ) = −i 3 2 To simplify higher powers i = i i = ( − 1)( − 1) = 1 i and see what is left. 4 2 2 of i then, we'll group all the 4ths i = i i = 1( i ) = i 5 4 i = ( i ) i = (1) i = i 33 4 8 8 i = i i = 1( − 1) = −1 6 4 2 4 will go into 33 8 times with 1 left. i = i i = 1( − i ) = −i 7 4 3 i = ( i ) i = (1) i = −i 4 20 3 83 20 3 i = i i = 1(1) = 1 8 4 4 4 will go into 83 20 times with 3 left.
7. 7. This "discriminates" or tells us what type of solutions we'll have. − b ± b − 4ac 2 ax + bx + c = 0 2 x= 2a If we have a quadratic equation and are considering solutions from the complex number system, using the quadratic formula, one of three things can happen. 1. The "stuff" under the square root can be positive and we'd get two unequal real solutions if b 2 − 4ac > 0 2. The "stuff" under the square root can be zero and we'd get one solution (called a repeated or double root because it would factor into two equal factors, each giving us = if b 2 − 4acthe0same solution). 3. The "stuff" under the square root can be negative and we'd get two complex solutions that are conjugates of each− 4ac < 0 if b 2 other. The "stuff" under the square root is called the discriminant.