# Complex numbers

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##### Full Member
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

##### Super Moderator
Staff member
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

##### Elite Member
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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• topsquark

##### Full Member
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

##### Elite Member
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

##### Elite Member
$$\displaystyle 0+\left(-\frac{1}{2}\right)i$$

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

##### Full Member
(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

• JeffM

#### mmm4444bot

##### Super Moderator
Staff member
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. • topsquark

#### MarkFL

##### Super Moderator
Staff member
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:

$$\displaystyle a+bi$$

• topsquark and mmm4444bot

#### Jomo

##### Elite Member
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

##### Elite Member
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

##### Elite Member
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.

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

##### Elite Member
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

##### Super Moderator
Staff member
The OP's question has been answered, and so I'm going to close this thread

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