Question

Hi friends here is the tutorial on Introduction to complex numbers

Answer 1

the general form of a complex number is a + bi ...

let me first give u the description of iota ( i ) :
Have you ever thought about \(\sqrt{-1} , \sqrt{-2} \) ?
These are refferred to as : Imaginary numbers that is who can only be imagined and who dont exist :

\[\huge{\sqrt{-1}=i}\]
iota has the value \(\sqrt{-1}\)

Complex numbers representation : Complex numbers are represented by Z
general form of Z = (a+bi)

Addition of complex numbers :
let a complex number \(Z_1 = a+bi \) and \(Z_2 = c+d i\)

\[\large{Z_1+Z_2 = a+bi + c+di = (a+c)+(bi+di) = (a+c)+i(b+c)}\]
Note : Real part of a complex number Z is represented by : Re(z)
and imaginary part of Z is represented by Im(z)

hence we can say that \(Z_1+Z_2=(Re(z_1)+Re(z_2))+i(Im(z_1)+Im(z_2))\)

Subtraction Of complex numbers :

let us take \(Z_1=a+bi\) and \(Z_2=c+di\) as two complex numbers :

\[\large{Z_1-Z_2=(a+bi)-(c+di)}\]
\[\large{Z_1-Z_2=(a-c)+i(b-d)}\]

we can write it as
\[\large{Z_1-Z_2=(Re(Z_1)-Re(Z_2))+i(Im(Z_1)-Im(Z_2))}\]

Multiplication of complex numbers :

we are taking same complex numbers here also :

\[\large{Z_1*Z_2=(a+bi)(c+di)}\]
\[\large{Z_1*Z_2=ac+adi+bci+bdi^2}\]
\[\large{Z_1*Z_2=(ac-bd)+i(ad+bc)}\]

You must have having one question here how did i write \(bdi^2=-bd\)
since \(i=\sqrt{-1}\) therefore we can say that \(i^2=(\sqrt{-1})^2\)
\(i^2=-1\)
hence \(\large{bdi^2=bd(-1)=-bd}\)

Divison of complex numbers

\[\large{\frac{Z_1}{Z_2}=\frac{(a+bi)}{(c+di)}}\]
\[\large{\frac{Z_1}{Z_2}=\frac{(a+bi)(c-di)}{(c+di)(c-di)}}\]
\[\large{\frac{Z_1}{Z_2}=\frac{(ac-adi+bci-bdi^2)}{(c^2-di^2)}}\]
\[\large{\frac{Z_1}{Z_2}=\frac{(ac+bd)-i(ad-bc)}{c^2+d^2}}\]

Answer 2

Hope this helps u all .. i will post some advanced level of complex numbers tutorial soon

Answer 3

i really love complex nos. really ..so,,a good job done :)

Answer 4

Can I ask a question?

Answer 5

yes ask

Answer 6

Anyway, I asked :P
Why is it like this:
sqrt{-25} x sqrt{-4}
≠ sqrt{-25 x -4}
≠ sqrt{100}
≠ 10

Answer 7

5i * 2i = 10i^2= - 10 remember ... \(\sqrt{a}*\sqrt{b}=\sqrt{ab}\) is only true for some cases
\[i* \sqrt{a}*i*\sqrt{b}=i^2\sqrt{ab}=-1\sqrt{ab}\]

Answer 8

Yes I understand that.
But why can't we do in that way....

Answer 9

u can not apply that formula here ... this can only be possible when a and b are +ve or only 1 either a and b are negative. .

Answer 10

And that's the question - why?

Answer 11

let me ask u this question to ask u this 1
:
n+1/n+2=0
find n +1=0 , n =-1 ? right ?

Answer 12

Yes

Answer 13

wait .... in that question
\[\sqrt{100}=\pm10\]

Answer 14

got it now ?

Answer 15

Hmm... not really...
why that + one doesn't work is the question.

Answer 16

um... wait...
\(\large \sqrt{100}=10 \)
\(\large -\sqrt{100}=-10 \)

Answer 17

Just a minor point: Typically lowercase \(z\) is used, not uppercase.

Answer 18

sorry @Limitless

@Callisto what is i * i ?

Answer 19

@mathslover, it's no problem. I just wanted to point that out--it's not a big deal at all. :P

Answer 20

thanks @Limitless btw how is the tutorial ?

Answer 21

I know how to solve it.
I'm asking why we can't multiply the - and - to make a positive.

Answer 22

If I turn it into i, I can solve it.

Answer 23

\(\sqrt{-1}*\sqrt{-1}=\sqrt{-1*-1}=\pm\sqrt{1}=\pm1\)

+ does not satisfy that equation as \((\sqrt{-1})^2=-1\)

Answer 24

really nice..........well done!

Answer 25

hence we have to think about the condition about that ....

Answer 26

thanks @nitz

Answer 27

And that's the problem why it doesn't satisfy when it is the ''correct'' way to do it?

Answer 28

as i said we have to just think about it practically ..

Answer 29

No. Shouldn't we understand the basic concept?
Anyone can have a chance to fall into this trap.

Answer 30

yes ..

Answer 31

http://www.jimloy.com/algebra/two.htm will this make any sense ?

Answer 32

"Imaginary numbers that is who can only be imagined and who dont exist"
Imaginary numbers exist just as much as real numbers.

\(i\) technically has two values, \(\sqrt{-1}\) and \(-\sqrt{-1}.\) (This is deduced from \(i^2=-1.\))

The real parts and imaginary parts are formally denoted \(\Re(z)\) and \(\Im(z)\).

Typically one expresses the division of two complex numbers as a sum of fractions as follows:
\[\frac{z_1}{z_2}=\frac{ac+bd}{c^2+d^2}+i\frac{bc-ad}{c^2+d^2}.\]

Answer 33

@mathslover How does it related to my question?

Answer 34

*relate

Answer 35

mathematics fallacies.. isn't that a fallacy that we are talking abt

Answer 36

@Callisto,
\[\sqrt{a}\sqrt{b}=\sqrt{ab} \Leftrightarrow a,b \ge 0\]
That is why your method does not work.

Answer 37

that i what i said yupii

Answer 38

............... So, everyone is just telling me that it must be positive but not explaining why

Answer 39

hate to spam but i need help

Answer 40

but i asked u why is i^2=-1 and not 1 do u have any answer ?

Answer 41

@yeababy np ask

Answer 42

okay it's -1 and then?

Answer 43

why it is not 1

Answer 44

so is i^4 = 1

Answer 45

I told you when I just make it into i form, then it's not a question to me anymore. But the first time I solved this question, I forgot to change it into i form and I got that wrong

Answer 46

\[\sqrt{-1} \ ^2 = -1\]

Answer 47

Campbell: Yes.
Callisto: The trick where you multiply radicands only technically works for positive numbers.

Answer 48

To be honest, \(\sqrt{a}\sqrt{b}=\sqrt{ab} \Leftrightarrow a,b \ge 0\) is a definition, I think. It's the definition because negatives impose the wrong result. For example, let \(a\) be negative and \(b\) be negative. Following this identity, we have \(\sqrt{a}\sqrt{b}=\sqrt{ab}=\sqrt{|a||b|}\).
However, \(\sqrt{a}\sqrt{b}=\sqrt{-|a|}\sqrt{-|b|}=i\sqrt{|a|}i\sqrt{|b}=i^2\sqrt{|a|}\sqrt{|b}=-\sqrt{|a|}\sqrt{|b|}\)
Notice that \(-\sqrt{|a|}\sqrt{|b}\ne \sqrt{|a||b|}\) for any \(a,b \in \mathbb{R}.\)

Answer 49

Correction: \[-\sqrt{|a|}\sqrt{|b|}\ne \sqrt{|a||b|}\]

Answer 50

well i think the tutorial that i wanted to show may be not attract as bcz of this disccusion jk :)

Answer 51

Well, it's just a problem I encountered on OS. I know how to get the right answer but don't understand why that was wrong. It's also weird for me that we see a wrong result, so we exclude it and call it as a definition. Honestly, it rather looks like a condition for me. I'm really sorry if the question raised makes your post dull.

Answer 52

Callisto: You must be clear about what context you are working in. You are used to working in the real numbers. In the real numbers, \(\sqrt{-4}\times\sqrt{-25}\) is undefined, because the square root operation is not defined for negative numbers. In the complex numbers, that expression is defined, and it simplifies to -10.

Answer 53

@nbouscal, I was also thinking of that perspective. I wasn't sure if we could argue the point from a field theory perspective.

Answer 54

Yes, obviously, I've forgotten the basic.

Answer 55

You are just using a trick that you learned that you can use, when in fact that trick does not always work. It is like learning to take a derivative by using certain tricks, and then not recognizing that in some cases the tricks don't work because the derivative doesn't exist.

Answer 56

Thanks! @Limitless and @nbouscal

Answer 57

You're welcome, @Callisto!

Answer 58

Top post @all

i think the problem with the square roots has something to do with