Complex Numbers What are complex numbers and how are they simplified?

Slide 2:

Complex number: any number in the form a + bi where
i =
Add and Subtract:
1. (2 + 7i) + (3 – 5i) 2. (6 – 3i) - (7 – 2i) Treat “i” as you would any variable.
Remember: You can only add or subtract “like terms” 5 – 2i (6 – 3i) + (-7 + 2i) Since it’s subtraction,
Change the signs in the
Second quantity -1 - i

Slide 3:

Multiply:
(8 + 2i)(7 – 3i) 2. (8 – i)(3 + 2i)
3. (5 + 4i)(5 – 4i) FOIL just as you would if it were “x” 56 – 24i + 14i – 6i2 I2 = -1 56 – 24i + 14i – 6(-1)
56 – 24i + 14i + 6 Combine like terms 62 – 10i 24 + 16i – 3i – 2i2
24 + 16i – 3i – 2(-1)
24 + 16i – 3i + 2 26 + 13i 25 – 20i + 20i – 16i2
25 – 20i + 20i – 16(-1)
25 – 20i + 20i + 16
41 Note the set up of this problem –
The “imaginary” term canceled itself!

Slide 4:

“Divide” -
You cannot have a radical in the denominator and
“I” is a radical! You need to rationalize the
Denominator!
1. In order to “eliminate” the i in the denominator, multiply by i. This must
be done to the numerator also so we are only multiplying by 1! Since i2 = -1…

Slide 5:

Let’s look at another!

Slide 6:

That process doesn’t work for this type though. Instead,
Remember the problem where the “I” canceled out? We
Need to multiply by the “conjugate” The “conjugate” is 4 + 2i Distribute the numerator
FOIL the denominator I2 = -1 Simplify

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By: rahulprati (47 month(s) ago)

plz let me download it......