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February 22, 2026, 03:54:31 am

Author Topic: Maths Questions ^_^  (Read 9862 times)  Share 

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dcc

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Re: Maths Questions ^_^
« Reply #30 on: April 28, 2008, 10:40:40 pm »
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Complex Numbers were created because mathematicians were getting annoyed at equations like (which of course has no solution in the reals).  They are an extension on the real numbers which includes a new unit called the 'imaginary unit'.  Mathematicians define this 'imaginary unit' as .

All complex numbers () take the form:

, where i is this imaginary unit i discussed earlier.

ok, the main operations to know are:
Real part of a Complex Number:
Say you had a complex number , the REAL component of this is the part of z which is not proceeded by an I, so you can say:


More generally: if , then



Imaginary Part of a Complex Number:
The imaginary component of a complex number is the bit which is proceeded by an i (which is why it is called the imaginary part):

If , then:



More generally: if , then:



Complex Conjugate:
The complex conjugate of a complex number is given by changing the sign of the imaginary part.  Graphically it represents a reflection in the axis (but thats on argand diagrams, so you might not of seen them).

If you have:
then the complex conjugate of z is:

(just change the negative to positive, or positive to negative)

Note that this little bar: indicates a complex conjugate and is achieved in latex by typing \overline{z}

Addition:
(i.e. to add two complex numbers, you add the real & imaginary bits of each number)

Subtraction:
(as you would expect)

Multiplication::
Multiplication is much like expanding brackets (it IS expanding brackets) but you have to remember to simplify numbers etc:


Division:
Division is weird, so it requires a bit of thought.  I'll give you an example just straight up:

Say you wanted to find:

The first step is to multiply both top and bottom lines by the complex conjugate of the bottom line.  This is done because if you multiply a complex number with its conjugate, you generally get a real number, which simplifies the process immensely.



In General:



That should be all you need for complex numbers.
« Last Edit: April 28, 2008, 10:45:52 pm by dcc »

Odette

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Re: Maths Questions ^_^
« Reply #31 on: April 28, 2008, 10:45:49 pm »
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Ok thanks Dcc, I think I get it now, so I'll have a go at the questions myself, then I'll post them up to see if I'm on the right track.

Odette

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Re: Maths Questions ^_^
« Reply #32 on: April 28, 2008, 11:40:13 pm »
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Ok here goes nothing lol....
1 a) (4 + 8i) (8 + 6i) = (4 8 - 8 6) + (4 6 + 8 8)i
                          = (32 - 48) + (24 + 64)i
                          = -16 + 88i
 
 b) (4 + 8i) + 3(8 - 6i)= (4 + 8) + 3(8 - 6)i
                              = 12 + 24i -18i
                              = 12 + 6i  (Not too sure about this one)  
                         
c) =
                                                  =
                                                  =
                                                  = + i

d) =
                                                              =
                                                              =
                                                              =
                                                              = 5 + 6i


2. (4 + 8i) = 41984 - 38912i (can't be bothered typing it all out for this one)

Please let me know if these are right or wrong, thanks :)
« Last Edit: April 29, 2008, 08:15:51 am by Odette »

Odette

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Re: Maths Questions ^_^
« Reply #33 on: April 29, 2008, 08:46:30 pm »
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Anyone? Please ^_^
Oh never mind then, i'll have to figure it out myself, thanks for the explanation dcc.
« Last Edit: April 29, 2008, 10:20:20 pm by Odette »

Odette

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Re: Maths Questions ^_^
« Reply #34 on: May 15, 2008, 06:47:31 pm »
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Ok i really need help with something... evaluate (4 + 8i) ^1/2 using de moivre's theorem, I have the answer but it's in a+bi form, I want the full answer but my calculator wont give me the answer >.< I have a TI-89, would anyone be able to tell me how to get it into decimal form, so that its just one answer rather than in a+bi form?
Thanks

AppleXY

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Re: Maths Questions ^_^
« Reply #35 on: May 15, 2008, 06:53:19 pm »
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Press the green diamond button and "ENTER" to obtain an approximate value for any calculation :)

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Odette

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Re: Maths Questions ^_^
« Reply #36 on: May 15, 2008, 06:55:10 pm »
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Thank you ^_^
It didn't work >.<
« Last Edit: May 15, 2008, 06:57:05 pm by Odette »

Odette

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Re: Maths Questions ^_^
« Reply #37 on: May 15, 2008, 07:09:18 pm »
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Oh and another question (8+6i) ^1/3 ... I keep getting the same answer and it's wrong.
Here's what i did:
r = sqrt (8^2 + 6^2)
r= sqrt (100)
r= 10

= tan^-1 (6/8)
 = 0.64

z^1/3= 10 ^1/3 (cos (1/3 * 0.64) + i sin (1/3 *0.64) )
        = 10 ^1/3 (cos (0.22) + i sin (0.22))
        = 2.15 e ^0.22i
        = 2.10 + 0.46i
« Last Edit: May 15, 2008, 07:11:30 pm by Odette »

dcc

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Re: Maths Questions ^_^
« Reply #38 on: May 15, 2008, 07:15:41 pm »
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Solutions for are:



But that is not a question which favours the use of de'moivre.
« Last Edit: May 15, 2008, 07:21:34 pm by dcc »

Odette

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Re: Maths Questions ^_^
« Reply #39 on: May 15, 2008, 07:20:39 pm »
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So my answer is wrong? -is confused-

Oh and another question, how exactly to you print a graph you've sketched using you calculator?
Sorry I don't use my calculator much.

Nevermind I used a graphing program ^_^
« Last Edit: May 15, 2008, 07:48:14 pm by Odette »

Mao

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Re: Maths Questions ^_^
« Reply #40 on: May 15, 2008, 08:25:59 pm »
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Oh and another question (8+6i) ^1/3 ... I keep getting the same answer and it's wrong.
Here's what i did:
r = sqrt (8^2 + 6^2)
r= sqrt (100)
r= 10

= tan^-1 (6/8)
 = 0.64

z^1/3= 10 ^1/3 (cos (1/3 * 0.64) + i sin (1/3 *0.64) )
        = 10 ^1/3 (cos (0.22) + i sin (0.22))
        = 2.15 e ^0.22i
        = 2.10 + 0.46i
why is this wrong?

there are two more solutions, is that what you meant?

remember:

, where k is an integer.
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Odette

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Re: Maths Questions ^_^
« Reply #41 on: May 15, 2008, 09:46:41 pm »
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Huh? What's k? I don't get it >.<

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Re: Maths Questions ^_^
« Reply #42 on: May 15, 2008, 11:22:26 pm »
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Complex Numbers were created because mathematicians were getting annoyed at equations like (which of course has no solution in the reals).  They are an extension on the real numbers which includes a new unit called the 'imaginary unit'.  Mathematicians define this 'imaginary unit' as .

Has anyone defined the log of a negative number? :D

Mao

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Re: Maths Questions ^_^
« Reply #43 on: May 15, 2008, 11:27:39 pm »
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:P
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AppleXY

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Re: Maths Questions ^_^
« Reply #44 on: May 16, 2008, 09:11:07 pm »
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Huh? What's k? I don't get it >.<

k is an integer, like 1,2,3...

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