Question

Can any one explain me complex number chapter plz...

Answer 1

plz

Answer 2

be back in 1 min wait.

Answer 3

0k waiting

Answer 4

anyone there ?

Answer 5

Yeah lets start

Answer 6

ok

Answer 7

\[i=\sqrt{-1}\]\[\omega+\omega^2+1=0\]

Answer 8

@soty2013 ...i Suggest u to View any Tutorial...on Youtube...May be Khan Academy...

Answer 9

^ Good idea.

Answer 10

@hba will you too suggest me the same ?

Answer 11

@soty2013

Let me help you a little and then you can leave and watch the khan academy video.

Answer 12

@soty2013

Yeah so tell me what is

\[\sqrt{16}=\]

Answer 13

4

Answer 14

What about,

\[\sqrt{-16}=\]

Answer 15

@soty2013

Yeah ?

Answer 16

4i

Answer 17

Yeah good. :)

Answer 18

When any number k is squared,the result is whether positive or zero i,e,\[k^2 \ge 0\]
Such numbers are called REAL numbers.On the other hand when \[k^2 < 0\] Such number are called IMAGINARY numbers

Example :

\[\sqrt{-1},\sqrt{-7},\sqrt{-20}\]

Answer 19

\[i=\sqrt{-1},where \ "i" \ represents \ ' i ota'\]\[i^2=-1\]\[i^3=(i^2)(i)=-i\]\[i^4=(i^2)(i^2)=1\]\[i^-1=\frac{ 1 }{ i }.\frac{ i }{ i }=\frac{ i }{ i^2 }=\frac{ i }{ -1 }=-i\]

Answer 20

ok

Answer 21

Complex numbers

When a real number and an imaginary number is added or subtracted the expression so formed,which cannot be simplified is called a complex number.

For Example,\[a+ib\]

Answer 22

OPERATIONS ON COMPLEX NUMBERS
-------------------------------

ADDITION
^^^^^^^

Real terms and imaginary terms are compunded in two seprate groups.
For example:(2+3i)+(3+5i)=(2+3)+(3i+5i)=5+8i

i.e (a,b)+(c,d)=(a+c,b+d)

Answer 23

ok

Answer 24

SUBTRACTION
^^^^^^^^^^

Real terms and imaginary terms are compunded in two seperate groups.
For example:(4+2i)-(2-5i)=(4-2)+[2i-(-5i)]=2+7i
i.e (a,b)-(c,d)=(a-c,b-d)

Answer 25

MULTIPLICATION
^^^^^^^^^^^^

The distributive law of multiplication applied to two complex numbers gives their product,

For example : (2+3i)(4-i)=8-2i+12i-3i^2=11+10i

Answer 26

i.e (a,b).(c,d)=(ac-bd,ad+bc)

Answer 27

http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/

Answer 28

DIVISION
^^^^^^

\[(1+i) \div (1-i) = \frac{ 1+i }{ 1-i } \times \frac{ 1+i }{ 1+i }=\frac{ (1+i^2) }{ (1)^2-(i)^2 }=i\]

i.e
\[(a,b) \div (c,d)=(\frac{ ac+bd }{ c^2+d^2 },\frac{ bc-ad }{ c^2+d^2 })\]

Answer 29

@soty2013: Pop quiz :D
Rationalize and "realize" the denominator:\[{1 \over \sqrt{6} +i}\]

Answer 30

COMPLEX CONJUGATE
^^^^^^^^^^^^^^^^
Any pair of complex numbers a+ib and a-ib have a product which is real,since (a+ib)(a-ib)=a^2-abi+abi-b^2i^2=a^2+b^2
Such complex numbers are said to be conjugate and each is the conjugate of the other

If a+ib is denoted by z,then its conjugate a-ib is denoted by z' .

Answer 31

Thanks @hba

Answer 32

I am not yet done.

Answer 33

EQUAL COMPLEX NUMBERS
-----------------------

Two complex numbers are equal if and only if the real terms and the imaginary terms are seprately equal

Answer 34

CUBE ROOT OF UNITY
^^^^^^^^^^^^^^^^

Consider the equation x^3-1=0

(x-1)(x^2+x+1)=0


\[x=1,x=\frac{ -1+i \sqrt{3} }{ 2 },x= \frac{- 1-i \sqrt{3} }{ 2 }\]