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HEY!! i need help with these problems! thank you!!!


Write each expression asa complex number in standard form:


(3+i)(4+i)


(3-4i)(4-3i)


(2+6i)(2-6i)


3+2i/3-2i


√6/√6-i




thank you so much!

=]

2007-10-14 11:08:26 · 8 answers · asked by bosstastic<3 3 in Science & Mathematics Mathematics

8 answers

1. 12 +3i +4i - 1 = 11+7i

2. 12 - 9i -16i - 12 = -25i

3. 4 - 12i +12i +36 = 40

4. [3+2i/3-2i][3+2i/3+2i]= (5+12i)/13

5.(√6/√6-i)(√6+i/√6+i)= (6 +√6i)/7

2007-10-14 11:35:10 · answer #1 · answered by Anonymous · 2 0

It's been a while for me since imaginary numbers but I'll give it a go. I'm going to assume that you know how to expand the product of to binomials using the FOIL method or some other way.
Remember:
i=√-1
i² = -1
i³ = -i
iΛ4 = 1

1. (3+i)(4+i)
12 + 7i + i²
12 + 7i - 1
11 + 7i

2. (3 - 4i)(4-3i)
12 - 25i + 12i²
12 - 25i - 12
-25i

3. (2+6i)(2-6i)
4 - 36i²
4 + 36
40

4. 3+2i/3-2i (I assume you mean (3+2i)/(3-2i).)
Multiply the denominator and the numerator by (3+2i).

9 + 12i + 4i²
-----------------
9 -4i²

5 + 12i
----------
13

5/13 + 12i/13

5.√6/√6-i
Multiply top and bottom by √(6+i)
You end up with:
√[(36+6i)/37]

Hope this helped and good luck!

2007-10-14 11:52:30 · answer #2 · answered by zook388 1 · 0 0

(3+i)(4+i)
= 12 + 7 i -1 = 11 + 7 i

(3 + 4 i ) (4 - 3 i)
= 12 + 7 i + 12
= y i

(2+ 6i) (2 - 6 i)
= 4 + 36
= 40

3 + 2 i / 3 - 2 i
= (9 + 2i - 6i ) / 3
= (9 - 4i ) / 3
= 3 - 4/3 i

1 - i

2007-10-14 23:17:32 · answer #3 · answered by sahil 2 · 0 0

(3+i)(4+i)
= 12 + 7 i -1 = 11 + 7 i

(3 + 4 i ) (4 - 3 i)
= 12 + 7 i + 12
= y i

(2+ 6i) (2 - 6 i)
= 4 + 36
= 40

3 + 2 i / 3 - 2 i
= (9 + 2i - 6i ) / 3
= (9 - 4i ) / 3
= 3 - 4/3 i

1 - i

2007-10-14 11:17:19 · answer #4 · answered by CPUcate 6 · 0 3

(3+i)(4+i) =
12 + 7i + i^2 =
11 + 7i

(3-4i)(4-3i) =
12 - 25i + 12i^2 =
-25i

(2+6i)(2-6i) =
4 - 36i^2 =
40

3+2i/3-2i =
(3+2i)(3+2i) / (3-2i)(3+2i) =
(9 + 12i + 4i^2) / (9 - 4i^2) =
(5 + 12i)/13

√6/√6-i =
√6*√(6+i) / √(6-i)*√(6+i) =
6 + i√6 / (6 -i^2) =
(6 + i√6) / 7

you're welcome

2007-10-14 11:21:45 · answer #5 · answered by Steve A 7 · 1 0

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2016-11-08 07:51:40 · answer #6 · answered by Anonymous · 0 0

Use the same methods that you would for all real numbers (FOIL'ing and the like). Since you're working with imaginary numbers, here's two things to remember:

i^2 = -1
1/(a+bi) = (a-bi)/(a^2 + b^2)

That should be all you need, so have at it!

2007-10-14 11:19:34 · answer #7 · answered by swrogueman 2 · 0 3

How 'bout you try paying attention in class so you'd know how to do your homework...

Just consider it....

2007-10-14 11:16:09 · answer #8 · answered by Anonymous · 0 5

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