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     CDSE

    Department of Engineering Mathematics

      Subject Code: BE 401 Subject: Engineering Mathematics –III

    utoria! opic

    1 Analytic Function, C-R Equations

    " Complex Integration

    # Poles , Residue, Singularity

    4$ Contour Integration

    utoria! – 1

    %eca!!: 1. Functions o complex !aria"le, its limit, continuity and dierentia"ility

    #. Analytic unctions and t$eir properties

    %. &ierence "et'een poles, residue and singularity

    (. Contour integration

    &1$ I),*   y xu

     and),*   y xv

      are $armonic unctions in a region R, pro!e t$at

       

      

     ∂∂

    +∂∂

    +   

      

     ∂∂

    −∂∂

     y

    v

     x

    ui

     x

    v

     y

    u

      is an analytic unction o  z = x+ i y .

    &$" Find t$e imaginary part o t$e analytic unction '$ose real part is

    % x  ### %%%   y x xy   −+−

    $ S$o' t$at t$e unction f  ( z )= xy+iy  is continuous e!ery'$ere "ut is not analytic.

    &$4 I)* z  f  

     is a regular unction o z 

     pro!e t$at

    #+#

    #

    #

    #

    #

    )*()*   z  f   z  f   y x

    =   

      

     ∂∂

    +∂∂

    &$' E!aluate ∫0

    2+i

    ( ´ z )2 dz ,  along

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    *i) $e real axis to # and t$en !ertically to #i.

    *ii) $e line y=

    1

    2 x

    &$( Find t$e order o eac$ pole and residues at it or t$e unction1

    1( + z 

    .

    &$):  S$o' t$at an analytic unction 'it$ constant modulus is constant

    &$* using Cauc$ys integral ormula, e!aluate ∫c

    ❑e2  z

    ( z+1 )4 dz

     , '$ere C is t$e circle I / I 0%.

    &$+ E!aluate ∫C 

    ❑cosπ z

    2

    ( z−1 )( z−2)dz

    , '$ere C is t$e circle I/I 0%.

    &$10 sing contour integration E!aluate

    #

    2  # cos

    d π θ

    + θ∫ 

    &$11 Pro!e t$at

    ( )11

    #

    cos#1

    #cos   ##

    ##

    2

    #  <−

    =+−∫ 

      aa

    ad 

    aa

    π θ 

    θ 

    θ π  

    ,ns-ers:

    3#.   3 x2

     y+6 xy− y3+c , '$ere c is a constant.

    3.4 *i)

    1

    3(14+11i)

      *ii)

    5

    3(2−i)

    3.5−14 (1+i√ 2 ) ,

    −14 (−1+i√ 2 ) ,

    −14 (−1−i√ 2 ) ,

    −1

    4 (1−i√ 2 )

    3.68π e

    −2

    3i

    3.7 (   πi

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    3.12

    2 π 

    √ 3