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Complex Analysis




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Complex Function
 Complex Exponential Function
 Properties

source/reference:
https://www.youtube.com/channel/UCaTLkDn9_1Wy5TRYfVULYUw/playlists

Complex Function

Complex Exponential Function

For the function, 𝑓(𝑧)=β„―π‘₯cos𝑦+𝑖ℯπ‘₯sin𝑦, (where 𝑧=π‘₯+𝑖𝑦) is an entire (=analytic in β„‚ function.

Some of its properties:

  • if 𝑦=0, then 𝑓(𝑧)=𝑓(π‘₯+𝑖⋅0)=𝑓(π‘₯)=β„―π‘₯, so 𝑓 agrees with the "regular" exponential function on ℝ
  • 𝑓(𝑧)=β„―π‘₯(cos𝑦+𝑖sin𝑦)=β„―π‘₯ℯ𝑖𝑦

By definition. The complex exponential function, ℯ𝑧, sometimes also denoted exp(𝑧), is defined by

ℯ𝑧=β„―π‘₯⋅ℯ𝑖𝑦, where 𝑧=π‘₯+𝑖𝑦

Properties

For the function, ℯ𝑧= β„―π‘₯⋅ℯ𝑖𝑦, where 𝑧=π‘₯+𝑖𝑦:

  • |ℯ𝑧|=|β„―π‘₯||ℯ𝑖𝑦|=β„―π‘₯
  • argℯ𝑧=arg(β„―π‘₯ℯ𝑖𝑦)=𝑦(+2πœ‹π‘˜, where π‘˜βˆˆβ„€)
  • ℯ𝑧+2πœ‹π‘–=β„―π‘₯ℯ𝑖(𝑦+2πœ‹)=β„―π‘₯ℯ𝑖𝑦=ℯ𝑧
  • ℯ𝑧+𝑀=β„―(π‘₯+𝑖𝑦)+(𝑒+𝑖𝑣), where 𝑧=π‘₯+𝑖𝑦, 𝑀=𝑒+𝑖𝑣  =β„―(π‘₯+𝑒)+𝑖(𝑦+𝑣)=β„―π‘₯ℯ𝑒ℯ𝑖𝑦ℯ𝑖𝑦  =(β„―π‘₯ℯ𝑖𝑦)(ℯ𝑒ℯ𝑖𝑦)=ℯ𝑧ℯ𝑀
  • 1𝑧=β„―βˆ’π‘§, since β„―π‘§β„―βˆ’π‘§=β„―0=1
  • ℯ𝑧 is an entire function.
  • Derivative 𝑓′(𝑧):

    Let 𝑒(π‘₯,𝑦)=β„―π‘₯cos𝑦, 𝑣(π‘₯,𝑦)=β„―π‘₯sin𝑦

    Then 𝑒π‘₯(π‘₯,𝑦)=𝑒π‘₯cos𝑦;𝑣π‘₯(π‘₯,𝑦)=𝑒π‘₯sin𝑦 𝑒𝑦(π‘₯,𝑦)=βˆ’π‘’π‘₯sin𝑦;𝑣𝑦(π‘₯,𝑦)=𝑒π‘₯cos𝑦

    Thus 𝑓′(𝑧)=𝑒(π‘₯,𝑦)+𝑖𝑣(π‘₯,𝑦)=β„―π‘₯cos𝑦+𝑖ℯπ‘₯sin𝑦=ℯ𝑧

    So the derivative of ℯ𝑧 is ℯ𝑧, in symbols, dd𝑧ℯ𝑧=ℯ𝑧.

  • ddπ‘§β„―π‘Žπ‘§=π‘Žβ‹…β„―π‘Žπ‘§ (π‘Žβˆˆβ„‚) by the chain rule
  • ℯ𝑧=β„―π‘₯βˆ’π‘–π‘¦=β„―π‘₯β„―βˆ’π‘–π‘¦=β„―π‘₯ℯ𝑖𝑦=β„―π‘₯ℯ𝑖𝑦=ℯ𝑧
  • ℯ𝑧=1 if and only if β„―π‘₯ℯ𝑖𝑦=1. The complex number in polar form, β„―π‘₯ℯ𝑖𝑦, equals 1, when its length equals 1 and its argument equals 0, ie.e. when β„―π‘₯ and y=2π‘˜πœ‹, π‘˜βˆˆβ„€. Thus

    ℯ𝑧=1⇔𝑧=2πœ‹π‘–π‘˜, π‘˜βˆˆβ„€
  • ℯ𝑧=β„―π‘€β‡”β„―π‘§βˆ’π‘€=1β‡”π‘§βˆ’π‘€=2πœ‹π‘–π‘˜β‡”π‘§=𝑀+2πœ‹π‘–π‘˜

    The function 𝑀=ℯ𝑧 is a mapping from β„‚ 𝑧-plane to β„‚ 𝑀-plane .

    For the images of horizontal lines, 𝐿={π‘₯+𝑖𝑦0|π‘₯βˆˆβ„} for fixed 𝑦0βˆˆβ„. Then ℯ𝑧=β„―π‘₯+𝑖𝑦0=β„―π‘₯ℯ𝑖𝑦0, a line from origin but not equal with fixed angle.

    For the images of vertical lines, 𝐿={π‘₯0+𝑖𝑦|π‘¦βˆˆβ„} for fixed π‘₯0βˆˆβ„. Then ℯ𝑧=β„―π‘₯0+𝑖𝑦=β„―π‘₯0ℯ𝑖𝑦, a circle with center at origin.

    For the images of vertical strip, 𝑆={𝑧:0<Re𝑧<1}, a ring between circle of value 0 and e

  • When ℯ𝑧=0
    ℯ𝑧=0⇔ℯπ‘₯⋅ℯ𝑖𝑦=0 Note: ℯ𝑖𝑦 has absolute value 1  β‡”β„―π‘₯=0  β‡”Never...!
  • For a given π‘§βˆˆβ„‚\{0}, is there a π‘€βˆˆβ„‚ such that ℯ𝑀=𝑧? Writing 𝑧=|𝑧|β„―π‘–πœƒ and 𝑀=𝑒+𝑖𝑣 this is equivalent to:
    ℯ𝑀=𝑧⇔ℯ𝑒ℯ𝑖𝑣=|𝑧|β„―π‘–πœƒ  β‡”β„―𝑒=|𝑧| and ℯ𝑖𝑣=β„―π‘–πœƒ  β‡”𝑒=ln|𝑧| and 𝑣=πœƒ+2π‘˜πœ‹  β‡”𝑀=ln|𝑧|+𝑖arg𝑧

    This is the complex logarithm.


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ID: 190400003 Last Updated: 2019/4/3 Revision:

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