Complex numbers techniques !

This note is for complex numbers lovers.

I have very keen interest in complex numbers and as many of you also are mad about using Complex Numbers in Various Section of Maths! Since It is Modern Mathematics Technique , And It Highly reduce our calculation .

So Here in this Note I want to share Some *Complex Number Techniques That I learnt till Now . And It is Humble Request to Post Related Complex number Techniques which You Had Learnt or Created at your Own ! So That Our Brilliant Community Learn from it.*


A few of them are :

\bullet 1 ) - Find Summation of Series :

C=cosθ+cos2θ+cos3θ+......+cos(nθ)S=sinθ+sin2θ+sin3θ+......+sin(nθ)C\quad =\quad \cos { \theta } +\cos { 2\theta } \quad +\cos { 3\theta } +\quad .\quad .\quad .\quad .\quad .\quad .\quad +\quad \cos { (n\theta ) } \\ \\ S\quad =\quad \sin { \theta } +\quad \sin { 2\theta } \quad +\sin { 3\theta } +\quad .\quad .\quad .\quad .\quad .\quad .\quad \quad +\quad \sin { (n\theta ) } .

By Using Euler's and Converting into :

C+iS=eiθ+ei2θ+ei3θ+.......ei(nθ)(G.P)C+iS=eiθ(ei(nθ)1)eiθ1C\quad +\quad iS\quad =\quad { e }^{ i\theta }\quad +\quad { e }^{ i2\theta }\quad +\quad { e }^{ i3\theta }\quad +\quad .\quad .\quad .\quad .\quad .\quad .\quad .\quad { e }^{ i(n\theta ) }\quad \quad \quad \quad (\quad G.P\quad )\\ \\ \\ C\quad +\quad iS\quad =\frac { { e }^{ i\theta }({ e }^{ i(n\theta ) }\quad -\quad 1) }{ { e }^{ i\theta }\quad -\quad 1 } .

And Then Separate real and imaginary Part And Then Compare !


\bullet 2 )- Find Integrals I1=eaxcosbxdx&I2=eaxsinbxdx{ I }_{ 1 }\quad =\int { { e }^{ ax }\cos { bx } } dx\quad \quad \& \quad { I }_{ 2 }\quad =\int { { e }^{ ax }\sin { bx } } dx\quad .

By Considering I1+iI1=eax(cosbx+isinbx)dx=e(a+ib)xdxI1+iI1=e(a+ib)xa+ib{ I }_{ 1 }\quad +\quad i{ \cdot I }_{ 1 }\quad =\int { { e }^{ ax }(\cos { bx } +i\cdot \sin { bx) } } dx\quad =\quad \int { { e }^{ (a+ib)x } } dx\quad \\ \\ { I }_{ 1 }\quad +\quad i{ \cdot I }_{ 1 }\quad =\quad \cfrac { { e }^{ (a+ib)x } }{ a+ib } .

And then Separate Real and imaginary Part and Then Compare !


\bullet 3 )- TPT : sinπ7sin2π7sin3π7=78\sin { \cfrac { \pi }{ 7 } } \cdot \sin { \cfrac { 2\pi }{ 7 } } \cdot \sin { \cfrac { 3\pi }{ 7 } } \quad =\quad \cfrac { \sqrt { 7 } }{ 8 } .

By Considering 7th Root's of unity :

1,α1,α2,.....,α6(αk=e2kπi7)1\quad ,\quad { \alpha }_{ 1 }\quad ,\quad { \alpha }_{ 2 }\quad ,\quad \quad .\quad .\quad .\quad .\quad .\quad ,\quad { \alpha }_{ 6 }\quad \quad (\because \quad { \alpha }_{ k }\quad =\quad { e }^{ \cfrac { 2k\pi i }{ 7 } }\quad ).

And Using Property n'th roots of unity ( Here n = 7 ):

(1α1)(1α2)(1α3).....(1α6)=7\left| (1\quad -\quad { \alpha }_{ 1 })(1\quad -\quad { \alpha }_{ 2 })(1\quad -\quad { \alpha }_{ 3 })\quad .\quad .\quad .\quad .\quad .\quad (1\quad -\quad { \alpha }_{ 6 }) \right| \quad =\quad \left| \quad 7\quad \right| .

And Now : 1αK=1(cos2πk7+isin2πk7)1αK=2sinπk7(cosπk7isinπk7)1αK=2sinπk7(II)(1α1)(1α2)(1α3).....(1α6)=726×(sinπ7sin2π7sin3π7)2=7sinπ7sin2π7sin3π7=781\quad -\quad { \alpha }_{ K }\quad =\quad 1\quad -\quad (\cos { \cfrac { 2\pi k }{ 7 } } \quad +\quad i\cdot \sin { \cfrac { 2\pi k }{ 7 } } )\\ \\ 1\quad -\quad { \alpha }_{ K }\quad =\quad 2\sin { \cfrac { \pi k }{ 7 } } (\cos { \cfrac { \pi k }{ 7 } } \quad -\quad i\cdot \sin { \cfrac { \pi k }{ 7 } } )\\ \\ \left| 1\quad -\quad { \alpha }_{ K } \right| =\quad 2\sin { \cfrac { \pi k }{ 7 } } \quad \quad \quad \quad \quad (II)\\ \\ \left| (1\quad -\quad { \alpha }_{ 1 })(1\quad -\quad { \alpha }_{ 2 })(1\quad -\quad { \alpha }_{ 3 })\quad .\quad .\quad .\quad .\quad .\quad (1\quad -\quad { \alpha }_{ 6 }) \right| =\quad 7\\ \\ { 2 }^{ 6 }\times { (\sin { \cfrac { \pi }{ 7 } } \cdot \sin { \cfrac { 2\pi }{ 7 } } \cdot \sin { \cfrac { 3\pi }{ 7 } } ) }^{ 2 }\quad =\quad 7\\ \\ \sin { \cfrac { \pi }{ 7 } } \cdot \sin { \cfrac { 2\pi }{ 7 } } \cdot \sin { \cfrac { 3\pi }{ 7 } } \quad =\quad \cfrac { \sqrt { 7 } }{ 8 } .


I'am Done ! Now It's Your Turn .

Please Post Your Complex Number Techniques Too !! :)

Enjoy Complex !! :) :)


Re-share This More And More So that it reaches to every complex numbers Lovers , So that we can learn new Techniques !

Note by Deepanshu Gupta
4 years, 11 months ago

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Expanding cos7x cos^{7}x in terms a series of cosines of multiples of x.

y=cosx+isinx y = cosx + isinx

yn+1yn=2cosnx y^{n} + \dfrac{1}{y^{n}} = 2cosnx , ( Since y=eixy={ e }^{ ix } )

y+1y=2cosxy + \dfrac{1}{y} = 2cosx

cos7x=(y+1y)7 cos^{7}x = ( y + \dfrac{1}{y})^{7}

=y7+7y5+21y3+35y+351y+211y3+71y5+1y7 = y^{7} + 7y^{5} + 21y^{3} + 35y + 35\dfrac{1}{y} + 21\dfrac{1}{y^{3}} + 7\dfrac{1}{y^{5}} + \dfrac{1}{y^{7}}

=(y7+1y7)+7(y5+1y5)+21(y3+1y3)+35(y+1y) = ( y^{7} + \dfrac{1}{y^{7}}) + 7(y^{5} + \dfrac{1}{y^{5}}) + 21(y^{3} + \dfrac{1}{y^{3}}) + 35(y + \dfrac{1}{y})

=2cos7x+7.2cos5x+21.2cos3x+35.2cosx = 2cos7x + 7.2cos5x + 21.2cos3x + 35.2cosx

cos7x=164(cos7x+7cos5x+21cos3x+35cosx) cos^{7}x = \dfrac{1}{64}(cos7x + 7cos5x + 21cos3x + 35cosx)

U Z - 4 years, 10 months ago

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limnr=1n(VeryNice)r\lim _{ n\rightarrow \infty }{ \prod _{ r=1 }^{ n }{ { (Very\quad Nice) }^{ r } } } .

Deepanshu Gupta - 4 years, 10 months ago

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Awsome @megh choksi thanks for sharing

Aman Sharma - 4 years, 10 months ago

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Well, there are a lot of applications of Euler's theorem.

For example :

1) r=0n(nr)cos(rx)=(r=0n(nr)eirx)\displaystyle \sum_{r=0}^{n} {n \choose r} \cos (r x) = \Re \bigg(\sum_{r=0}^{n} {n \choose r} e^{i r x}\bigg)

2) r=0cosrx2r=(r=0(eix2)r)\displaystyle \sum_{r=0}^{\infty} \dfrac{\cos rx}{2^r} = \Re \bigg(\sum_{r=0}^{\infty} \bigg(\dfrac {e^{ix}}{2} \bigg)^r \bigg)

3) Expressing sinrx\sin rx and cosrx\cos rx in terms of sinx\sin x and cosx\cos x

jatin yadav - 4 years, 11 months ago

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Did a silly thing , sorry i am posting it here

U Z - 4 years, 10 months ago

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Lol ! :)

You r creative :) :)

Deepanshu Gupta - 4 years, 10 months ago

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@Deepanshu Gupta awsone post............i need your help how to solve following problems:-

(1)..Let z1,z2,z3z_1,z_2,z_3 be three complex numbers such that:- z1=z2=z3=1z1+1z2+1z3=1|z_1|=|z_2|=|z_3|=|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|=1 Find the value of z1z2z3z_1z_2z_3

(2)let z be a complex number such that:- z+4z=2|z+\frac{4}{z}|=2 Find maximum value of |z|

(3)let z be a complex number then find maximum value of |z|+|z-1|

(4)if z2+z+1=0z^2+z+1=0 where z is a complex number then find tje value of:- (z+1z)2+(z2+1z2)2+...........+(z6+1z6)2(z+\frac{1}{z})^2 + (z^2+\frac{1}{z^2})^2+...........+(z^6+\frac{1}{z^6})^2

Please help me how to solve these problems...i know it is off topic i am sorry for that......i am a beginer so please post solutions...... also sorry to desturb you

Aman Sharma - 4 years, 11 months ago

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Hints:

Sol 4 ) - Note Roots of quadratic are ω,ω2Cuberootofunity\quad \omega \quad ,\quad { \omega }^{ 2 }\quad \longrightarrow \quad Cube\quad root\quad of\quad unity

And Use Following Properties To get Answer :

ω+ω2+ω3=0(ω+ω2=1)&ω=1ω2\quad \omega \quad +\quad { \omega }^{ 2 }\quad +\quad { \omega }^{ 3 }\quad =\quad 0\quad \quad \quad (\therefore \quad \omega \quad +\quad { \omega }^{ 2 }\quad =\quad -1\quad )\\ \quad \quad \quad \quad \quad \& \quad \quad \omega \quad =\quad \cfrac { 1 }{ { \omega }^{ 2 } } \quad .

Sol 2)- Let z=r\left| z \right| \quad =\quad r.

use Triangle inequality :

r4rz+4zr+4rr4r2...(I)r+4r2(UselessTruerbyAMGM)\\ \left| r-\cfrac { 4 }{ r } \right| \quad \le \quad \left| z\quad +\quad \cfrac { 4 }{ z } \right| \quad \le \quad r\quad +\quad \cfrac { 4 }{ r } \\ \\ \left| r-\cfrac { 4 }{ r } \right| \quad \le \quad 2\quad \quad \quad .\quad .\quad .(I)\\ \quad r\quad +\quad \cfrac { 4 }{ r } \quad \ge \quad 2\quad \quad \quad \quad (Useless\quad \because \quad True\quad \forall \quad r\quad \quad by\quad AM-GM).

Now Solve II equation and get required maximum value of "r" .

I will Post rest of two later !

Deepanshu Gupta - 4 years, 11 months ago

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3) I think there is something wrong with it as when a,ba,b increases where z=a+ibz=a+ib a,ba,b belongs to R, then |z| increases and also |z-1| so it will grow to infinity.

So the answer is infinity or a typo.

Gautam Sharma - 4 years, 10 months ago

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Just one word; Brilliant! :)

sanat mishra - 3 years, 11 months ago

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