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(-3+i) + (4+6i) =


(-4 - 6i) + (-2 - 3i)


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


(-2 + 5i) + (-4 - 2i)


(4 + 2i)(3 - i)


(-3 + i)(2 + 7i)


(5 + 2i)(5 - 2i)


4+2i over 3-i


3+i over 1-4i


-2-2i over 5-2i

2007-07-29 15:42:11 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

Hi,

Operations on Complex Numbers?

To add or subtract, combine like terms:

(-3+i) + (4+6i) = 1 + 7i <== answer

(-4 - 6i) + (-2 - 3i) = -6 - 9i <== answer

(4 + i) - (-3 + 2i) = 7 - i <== answer

(-2 + 5i) + (-4 - 2i)= -6 + 3i <== answer


Multiply like regular variables, but replace i² with -1 and simplify.

(4 + 2i)(3 - i) = `12 - 4i + 6i - 2i² = 12 + 2i - 2(-1) =
12 + 2i + 2 = 14 + 2i <== answer

(-3 + i)(2 + 7i) = -6 - 21i + 2i + 7i² = -6 - 19i + 7(-1) =
-6 - 19i - 7 = -13 - 19i <== answer

(5 + 2i)(5 - 2i) = 25 - 10i + 10i - 4i² = 25 - 4(-1) = 25 + 4 =29

29 <== answer

Multiply numerator and denominator both by the conjugate of the denominator.

....4+2i
...--------
....3-i

Multiply top and bottom by (3 + i).

...(4+2i)(3 + i)
...----------------
...(3-i)(3 + i)

12 + 4i + 6i + 2i²
----------------------
9 - i²

12 + 10i + 2(-1)
----------------------
9 - (-1)

12 + 10i - 2
----------------
9 + 1



10 + 10i
----------- = 1 + i <== answer
10




3+i
-----
1-4i

Multiply top and bottom both by denominator's conjugate,
(1 + 4i)

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

3 + 4i + i + 4i²
-------------------
1 - 16i²

3 + 5i + 4(-1)
-------------------
1 - 16(-1)


3 + 5i - 4
-------------
1 + 16

-1 + 5i
---------- <== answer
17



-2-2i
-------
5-2i

Multiply top and bottom both by denominator's conjugate,
(5 + 2i)

(-2-2i)(5 + 2i)
------------------
(5-2i)(5 + 2i)

-10 - 4i - 10i - 4i²
-----------------------
25 - 4i²

-10 - 4i - 10i - 4(-1)
-----------------------
25 - 4(-1)

-10 - 14i + 4
-----------------
25 + 4

-6 - 14i
---------- <== answer
29


I hope that helps!! :-)

2007-07-29 16:03:28 · answer #1 · answered by Pi R Squared 7 · 0 0

Hmmm... Why don't I show you how to do them eh?

Alright for addition and subtraction, what you do is treat the i like a variable and add and subtract accordingly:

(-3+i) + (4+6i) = 1+7i


For multiplication, again treat i as a variable, but now, FOIL it:

(4+2i)*(3-i) =
12 - 4i + 6i +2 = >>> Remember that i * i = -1
14 + 2i

For division, you have to know about complex conjugates, which is a big term for switch the middle sign.

So 3+i >>> 3-i

What you do is multiply by the complex conjugate of the denominator:

(4+2i / 3-i) * (3+i / 3+i)

(4+2i)(3+i) / (3-i)(3+i)

(10+10i) / (9+1)

(10 + 10i) / 10

1+i

There, all the rest are the same.

2007-07-29 22:51:53 · answer #2 · answered by AibohphobiA 4 · 0 0

These problems are easy, why do you want us to do them? The only ones that are slightly complicated are the divisions, but you just do those by multiplying numerator and denominator by the complex conjugate of the denominator, so that you end up with something real on the bottom.

2007-07-29 22:47:19 · answer #3 · answered by pegminer 7 · 0 0

I've made an app for this. Check it out below. It's easy to use and color coded for easy reading. This one adds and subtracts only.
http://www.tomsmath.com/adds-and-subtracts-complex-numbers.html

2014-06-25 19:24:57 · answer #4 · answered by ? 3 · 0 0

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