complex numbers?

Question

please!

#1
(8 - i) - (3+2i) + (-4 + 6i) write in standard form
answer?

#2
(2-3i)(-1+4i)

#3
3+2i/2+i

#4 find absolute value of 4+2i, 8i, and -3 + 7i.
which is farthest from origin?

Answer 1

Question 1
(8 - 3 - 4) + (- 1 - 2 + 6) i
1 + 3 i

Question 2
- 2 + 8 i + 3 i - 12 i ²
- 2 + 11 i + 12
10 + 11 i

Question 3
(3 + 2 i)(2 - i) / (2 + i)(2 - i)
(6 + i + 2) / 5
(8 + i) / 5

Question 4
| a | = | 4 + 2 i | = √20 = 2√5
| b | = | 8 i | = 8
| c | = | - 3 + 7 i | = √58

| c | is greatest so c is furthest from origin.

Answer 2

#1 Just treat them like polynomials: remove the brackets and collect like terms.

8 - i - 3 - 2i - 4 + 6i = 3i +1

#2 Use foil and simplify.

-2 + 8i + 3i - 12i^2 = -2 + 11i - 12i^2 (since i^2 = -1) =
-2 + 11i - 12(-1) = -2 + 11i + 12 = 10 + 11i

#3 multiply the numerator and denominator by 2 - i

[(3 + 2i) * (2 - i)] / (2 + i)(2 - i)] Please take it from here, you can do it :-)

#4 Absolute value of a complex number a + bi = sqrt(a^2 + b^2).

Again, take it from here and you will surely be able to say which is farthest. The larger the value the farther it is from the origin as the absolute value of the complex number is equal to the length of vector from the origin to the point (a, b)

Good Luck

Answer 3

1. 9-9i

2. 10+11i

3. 1.6+0.2i

4. 4.4721, 8, 7.62
8i is the furthest from the origin