perform each of the following calculations involving complex numbers, and write the answer, simplified as much as possible, in the form a + bi, where a and b are real numbers.

1. (4+2i) – (6 – 8i) (work not required)

2. i18 (work not required)

3. (1-3i)/(2+ i) Show work.

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1. (4+2i) - (6-8i):

distribute the minus sign:

= 4 + 2i - 6 + 8i

gather like terms:

= -2 + 10i

2. i18 = i times 18. You should write it as 0 + 18i to give the answer in the form asked for.

3. (1-3i) / (2+i):

here you have to multiply the top AND bottom by the "complex conjugate" of the bottom - much in the same way you simplify radical expressions to get rid of radicals in the denominator.

the complex conjugate of any complex number, a + bi, is a - bi

So you multiply the given expression by (2-i ) / (2 - i)

The numerator becomes (1 - 3i)(2 - i) which you can foil, or use the distribute law, or the "claw method" - whatever is easiest for you since these are all really the same method:

(1 - 3i)(2 - i) = 2 - 6i - i + 3i^2 = 2 - 7i +3(-1) = 2 - 7i - 3 = -1 - 7i

(remember i^2 = -1 )

The denominator becomes (2+i)(2-i) = 4 + 2i - 2i - i^2 = 4 - (-1) = 4 + 1 = 5

ANSWER: (-1 - 7i) / 5 which can be written in a + bi form as

- 1/5 - (7/5) i

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## Comments

^{what if it is i^18 }This is easy to remember because you already know the first 2 in this series (i^0 and i^1) and the next two or three make sense or are easy to figure out.

i^2 = -1

i^3 = -i

i^4 = 1

i^6 =i^2 = -1

i^7 =i^3 = -i

i^8 =i^4 = 1