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Function in W plane gets mapped to the Z plane as?

  1. Feb 5, 2017 #1
    1. The problem statement, all variables and given/known data
    A function is given as W = ln (Z)
    Z = x + i y
    W = u + i v
    i is sqrt of (-1)
    The u = constant lines get mapped in z plane as ?

    2. Relevant equations
    Z = x + i y = [ sqrt {x2 + y2 } ] ei (theta)
    where theta is tan-1 (y/x)

    3. The attempt at a solution
    Function w is
    w = ln (z)
    = ln( sqrt {[x2 + y2 ] } multiply with ei (theta)

    Now ln (a b) = ln (a) + ln (b)

    so w = ln (sqrt [x2 + y2 ]) + i(theta) ln (e)
    The first part is real, while second part is imaginary.
    w = u + iv.
    So u = ln(sqrt [x2 + y2 ] )
    Question says u = constant lines get mapped in z plane as ?
    u = constant
    so ln (sqrt (x2 + y2 ) = constant.

    Now how to proceed? In answer they've said concentric circles with radius ec.

    I got till u = ln (sqrt (x2 + y2)
    remove sqrt out as 1/2 we get:
    u = 1/2 ( ln (x2 + y2) )
    equation of circle is x2 + y2 = constant.

    But there is log in u.
     
  2. jcsd
  3. Feb 5, 2017 #1

    pasmith

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    Homework Helper

    Do you not recognize [itex]\ln\sqrt{x^2 + y^2} =\ln |z| = c[/itex] as [itex]|z| = e^c[/itex], which is a circle of radius [itex]e^c[/itex]?
     
  4. jcsd
  5. Feb 5, 2017 #2
    But question asks for mapping of lines u = constant.
    And not mapping of lines eu = constant.

    I understand how you removed log part by using exponential, but we need u = constant,
    and not eu = e constant

    Like left hand side is u and not eu.
     
  6. Feb 5, 2017 #3
    Ok now i get it. Mapping on z plane, means we need 'x' in terms of 'y'
    Equation that we get is: ln sqrt (x2 + y2 ) = constant c
    to get x in terms of y,
    sqrt (x2 + y2 ) = ec
    so squaring we get
    x2 + y2 = ec(2) --- equation of circle. radius ec
    that is x2 = e2c - y2

    Thanks.

    PS how do you put the sign of square root over x2 + y2
     
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