- Thread starter Ryan$
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\(\displaystyle a+bi\)

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 ..

\(\displaystyle a+bi\)

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:

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

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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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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?!

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

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

\(\displaystyle a+bi\)

Is this correct? (Please answer this question)

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

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

no didn't been asked, while solving a question I was encountered to convert that horrible number of complex to a+biI would guess [you] have been asked to write this number in the rectangular form:

\(\displaystyle a+bi\)

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\(\displaystyle \begin{align*}\dfrac{3-4\mathcal{i}}{7+5\mathcal{i}}&=\dfrac{(3-4\mathcal{i})(7-5\mathcal{i})}{49+25} \\&=\dfrac{(21-20)+\mathcal{i}(-15-28)}{74}\\&=\dfrac{1}{74}+\bigg( \dfrac{-43}{74} \bigg) \mathcal{i}\end{align*}\)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

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

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

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

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Regarding post #10:

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

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

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

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

… it is common to write … \(\displaystyle \tfrac{1}{74} - \tfrac{43}{74}i\)

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

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

\(\displaystyle a+bi\)

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Regarding post #10:

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

Correct & thank you.For the sake of the OP, I wanted to give the number strictly in the form: \(\displaystyle a+bi\)

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 \(\displaystyle a\)

If the one doing the correcting has a background in teaching

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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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:

\(\displaystyle a+bi\)

Let's see. You appear to accept an alleged "correction," but in the same motionCorrect & 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 \(\displaystyle a\)a real number \(\displaystyle b\) times \(\displaystyle \mathcal{i}\).plus

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.

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

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To Lookagain, where did I say anything about you? I did not. Did I?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.

But if the shoe fits then wear it.

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