Complex numbers

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Ryan$

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Hi guys,
the definition of complex numbers are in this form a+b*i
my question then why 1/(2*i) is complex number? the "i" isn't on the nominator .. then why we can say it's complex?!
 

MarkFL

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The form:

\(\displaystyle a+bi\)

is referred to as the rectangular form for a complex number, which is just one of many forms in which complex numbers may be written. It is a particularly useful form though. Also, observe:

\(\displaystyle \frac{1}{2i}=\frac{1}{2i}\cdot\frac{i}{i}=\frac{i}{2i^2}=-\frac{i}{2}=0+\left(-\frac{1}{2}\right)i\)
 

Ryan$

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The form:

\(\displaystyle a+bi\)

is referred to as the rectangular form for a complex number, which is just one of many forms in which complex numbers may be written. It is a particularly useful form though. Also, observe:

\(\displaystyle \frac{1}{2i}=\frac{1}{2i}\cdot\frac{i}{i}=\frac{i}{2i^2}=-\frac{i}{2}=0+\left(-\frac{1}{2}\right)i\)
I'm totally with you .. but once you've multiplied by i you've changed the cosmetics of the number .. isn't it matter? I mean isn't it matter for changing the definition of the number itself? who said that multiplying isn't changing the typecast of the number itself? really weird ..
 

MarkFL

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I've multiplied by 1 in the form of i/i, and thus have not changed the value of the number. The form is different, but it still represents the same complex value.
 

HallsofIvy

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The "cosmetics" of the number (I like that phrase!) is completely irrelevant to the problem. 3, III, 2+ 1, 6/2, and \(\displaystyle \sqrt{9}\) are all different ways of saying exactly the same thing. It does not matter to the mathematics of the problem which you use.
 
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Ryan$

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Hi guys !
lemme explain myself please and sorry if it's silly but I really struggle that.

lets assume I have complex number like (3-4i / 5j +7 ) then it's right to write like this :
(3-4i / 5j +7 ) = a + b*i

I write that but I don't know why ... I know that the complex number its form or standard form is a + b*i ! but what's confusing me how we are really equal something in general like a + b*i to specific thing like (3-4i / 5j +7 ) ! here's my problem .. any help?
when I put "=" then we say that (3-4i / 5j +7 ) is equal to a+b*i but a+b*i is general and (3-4i / 5j +7 ) is specific ... how that equal?!

Maybe please when you guys write that's (3-4i / 5j +7 ) = a + b*i , what you think about it?! or what makes you to believe that's right? general=specific ..don't make sense?!
 

MarkFL

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I would guess (I recommend you learn to use \(\LaTeX\) to make your expressions clear) that you have been given the complex number:

\(\displaystyle \frac{3-4i}{7+5i}\)

And have been asked to write this number in the rectangular form:

\(\displaystyle a+bi\)

Is this correct? (Please answer this question)

Note: I have seen both \(i\) and \(j\) used to represent the imaginary \(\sqrt{-1}\), but, I've never seen them both used in the same expression.
 

JeffM

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Ryan

You jump around from wondering whether 7 - 5 = 2 because 7 - 6 = 1 and 6 - 5 = 1, to questions about what the derivative is, to questions about complex numbers in a matter of days. This is ridiculous. Math is progressive and takes time to learn. Stick with one thing and LEARN IT before going off to something new.

I want to know the rectangular form of a specific complex fraction so I go

\(\displaystyle \dfrac{3 - 4i}{7 + 5i} = a + bi, \text { where } a \text { and } b \text { are UNKNOWNS.}\)

Letters in math can be used for different things in different problems. In this problem, a and b are not being used as parameters standing for any real numbers, but are standing for specific numbers. You must always identify what your letters mean relative to a specific problem.
 

Ryan$

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I would guess [you] have been asked to write this number in the rectangular form:

\(\displaystyle a+bi\)
no didn't been asked, while solving a question I was encountered to convert that horrible number of complex to a+bi
 

pka

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lets assume I have complex number like (3-4i / 5j +7 ) then it's right to write like this :
(3-4i / 5j +7 ) = a + b*i
\(\displaystyle \begin{align*}\dfrac{3-4\mathcal{i}}{7+5\mathcal{i}}&=\dfrac{(3-4\mathcal{i})(7-5\mathcal{i})}{49+25} \\&=\dfrac{(21-20)+\mathcal{i}(-15-28)}{74}\\&=\dfrac{1}{74}+\bigg( \dfrac{-43}{74} \bigg) \mathcal{i}\end{align*}\)

Thus \(\displaystyle a=\dfrac{1}{74}~\&~b=\dfrac{-43}{74}\) SEE HERE
 

lookagain

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\(\displaystyle 0+\left(-\frac{1}{2}\right)i\)

It is common to write this final result in "a + bi" form as \(\displaystyle \ 0 - \tfrac{1}{2}i\).

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Regarding post #10:

Likewise, it is common to write this final result in "a + bi" form as \(\displaystyle \ \tfrac{1}{74} - \tfrac{43}{74}i\).
 

topsquark

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(3-4i / 5j +7 ) = a + b*i
I would have called this a typo except that it has been carried through on a couple occasions.

If you are a Mathematician (or Physicist) \(\displaystyle i^2 = -1\) is common. If you are an Engineer then \(\displaystyle j^2 = -1\). You shouldn't mix the two. So your expression should either be
\(\displaystyle \dfrac{3 - 4i}{7 + 5i}\)
or
\(\displaystyle \dfrac{3 - 4j}{7 + 5j}\)

-Dan
 

mmm4444bot

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It is common to write … \(\displaystyle 0 - \tfrac{1}{2}i\)

… it is common to write … \(\displaystyle \tfrac{1}{74} - \tfrac{43}{74}i\)
Yes, in general, it's common to express the addition of a negative number as subtraction of its opposite. In particular, it's also common for authors who want to explicitly match the given form a+bi to stick with addition when b is negative by putting grouping symbols around it (like in posts #2 and #10).

Likewise, if emphasizing the given form when b is zero and a is not, authors will write a+(0)i instead of a.

😎
 

MarkFL

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It is common to write this final result in "a + bi" form as \(\displaystyle \ 0 - \tfrac{1}{2}i\).
For the sake of the OP, I wanted to give the number strictly in the form:

\(\displaystyle a+bi\)
 

Jomo

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I have always wondered about this a+bi form.
Because I am bothered by it I always say a +/- bi.
Now here is my question. If you are asked to write something in a+bi form is an answer of 4 - 5i in the wrong form? I think it is!
 

pka

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Regarding post #10:
Likewise, it is common to write this final result in "a + bi" form as \(\displaystyle \ \tfrac{1}{74} - \tfrac{43}{74}i\).
For the sake of the OP, I wanted to give the number strictly in the form: \(\displaystyle a+bi\)
Correct & thank you.
I have never figured why some want-to-bes must correct every little thing.
But in this case the "correction" is a mistake. It does say a real number \(\displaystyle a\) plus a real number \(\displaystyle b\) times \(\displaystyle \mathcal{i}\).
If the one doing the correcting has a background in teaching complex variables, it is standard to define the complex number field as a subset of \(\displaystyle \mathcal{R}\times\mathcal{R}\). As a field there are two operations \(\displaystyle {\bf\large+~\&~\cdot}\): \(\displaystyle (a,b)+(c,d)=(a+c,b+d)~\&~(a,b)\cdot(c,d)=(ac-bd,ad+bc)\).
Here is a basic idea: define a conjugate of a complex number, \(\displaystyle \overline{(a,b)}=(a,-b)\).
The absolute value is \(\displaystyle |(a,b)|=\sqrt{a^2+b^2}\) So the multiplicative inverse \(\displaystyle (a,b)^{-1}=\dfrac{\overline{(a,b)}}{|(a,b|^2}\)
Here is short-hand notation: \(\displaystyle \dfrac{1}{z}=\dfrac{\overline{z}}{|z|^2}\)
Apply this to the problem at the hand:
\(\displaystyle \dfrac{3-4\mathcal{i}}{7+5\mathcal{i}}=\dfrac{(3-4\mathcal{i})(7-5\mathcal{i})}{49+25}\)
 
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lookagain

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That top right denominator should be "49 + 25."

Likewise, it is common to write this final result in "a + bi" form as \(\displaystyle \ \tfrac{1}{74} - \tfrac{43}{74}i\).
For the sake of the OP, I wanted to give the number strictly in the form:

\(\displaystyle a+bi\)
Correct & thank you.
I have never figured why some want-to-bes must correct every little thing.
But in this case the "correction" is a mistake. It does say a real number \(\displaystyle a\) plus a real number \(\displaystyle b\) times \(\displaystyle \mathcal{i}\).
Let's see. You appear to accept an alleged "correction," but in the same motion
you dismiss it as petty. You are self-contradictory.
I'm not a wan't-to-be, but you sure are desperate and looking to remain relevant. When I stated
"it is common," that was not a correction. But the b can be negative at times,
so those forms of mine follow. Your math you posted is irrelevant/unneeded.


Jomo, stay with 4 - 5i for the form of your answer for your example.

It is readily seen that a = 4 and b = -5.

pka, this is the second post of yours that has been reported in recent weeks for you attacking me with your insults. You need to stop posting until you can be civil.
 
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pka

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pka, this is the second post of yours that has been reported in recent weeks for you attacking me with your insults. You need to stop posting until you can be civil.
To Lookagain, where did I say anything about you? I did not. Did I? Can you point to anything that I posted which namers you?
But if the shoe fits then wear it.
 

MarkFL

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The OP's question has been answered, and so I'm going to close this thread
 
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