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Complex Function βComplex Exponential Function βProperties
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Complex Function
Complex Exponential Function
For the function, π(π§)=β―π₯cosπ¦+πβ―π₯sinπ¦, (where π§=π₯+ππ¦) is an entire (=analytic in β function.
Some of its properties:
- if π¦=0, then π(π§)=π(π₯+πβ
0)=π(π₯)=β―π₯, so π agrees with the "regular" exponential function on β
- π(π§)=β―π₯(
cosπ¦+πsinπ¦)=β―π₯β―ππ¦
By definition. The complex exponential function, β―π§, sometimes also denoted exp(π§),
is defined by
β―π§=β―π₯β
β―ππ¦, where π§=π₯+ππ¦
Properties
For the function, β―π§= β―π₯β
β―ππ¦, where π§=π₯+ππ¦:
- |β―π§|=|β―π₯||β―ππ¦|=β―π₯
argβ―π§=arg(β―π₯β―ππ¦)=π¦(+2ππ, where πββ€)
- β―π§+2ππ=β―π₯β―π(π¦+2π)=β―π₯β―ππ¦=β―π§
β―π§+π€=β―(π₯+ππ¦)+(π’+ππ£), where π§=π₯+ππ¦, π€=π’+ππ£
=β―(π₯+π’)+π(π¦+π£)=β―π₯β―π’β―ππ¦β―ππ¦
=(β―π₯β―ππ¦)(β―π’β―ππ¦)=β―π§β―π€
1ℯπ§ =β―βπ§, since β―π§β―βπ§=β―0=1
- β―π§ is an entire function.
- Derivative πβ²(π§):
Let π’(π₯,π¦)=β―π₯cosπ¦, π£(π₯,π¦)=β―π₯sinπ¦
Then π’π₯(π₯,π¦)=ππ₯cosπ¦ ;π£π₯(π₯,π¦)=ππ₯sinπ¦
π’π¦(π₯,π¦)=βππ₯sinπ¦ ;π£π¦(π₯,π¦)=ππ₯cosπ¦
Thus πβ²(π§)=π’(π₯,π¦)+ππ£(π₯,π¦)=β―π₯cosπ¦+πβ―π₯sinπ¦=β―π§
So the derivative of β―π§ is β―π§, in symbols, ddπ§ β―π§=β―π§ .
ddπ§ β―ππ§=πβ
β―ππ§ (πββ) by the chain rule
- β―π§=β―π₯βππ¦=β―π₯β―βππ¦=β―π₯β―ππ¦=β―π₯β―ππ¦=β―π§
β―π§=1 if and only if β―π₯β―ππ¦=1. The complex number in polar form,
β―π₯β―ππ¦, equals 1, when its length equals 1 and its argument equals 0, ie.e.
when β―π₯ and y=2ππ, πββ€. Thus
β―π§=1βπ§=2πππ, πββ€
- β―π§=β―π€ββ―π§βπ€=1βπ§βπ€=2πππβπ§=π€+2πππ
The function π€=β―π§ is a mapping from β π§-plane to β π€-plane .
For the images of horizontal lines, πΏ={π₯+ππ¦0|π₯ββ} for fixed π¦0ββ. Then
β―π§=β―π₯+ππ¦0=β―π₯β―ππ¦0, a line from origin but not equal with fixed angle.
For the images of vertical lines, πΏ={π₯0+ππ¦|π¦ββ} for fixed π₯0ββ. Then
β―π§=β―π₯0+ππ¦=β―π₯0β―ππ¦, a circle with center at origin.
For the images of vertical strip, π={π§:0<Reπ§<1}, a ring between circle of value 0 and e
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When β―π§=0
β―π§=0ββ―π₯β
β―ππ¦=0 Note: β―ππ¦ has absolute value 1
ββ―π₯=0
βNever...!
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For a given π§ββ\{0}, is there a π€ββ such that β―π€=π§? Writing
π§=|π§|β―ππ and π€=π’+ππ£ this is equivalent to:
β―π€=π§ββ―π’β―ππ£=|π§|β―ππ
ββ―π’=|π§| and β―ππ£=β―ππ
βπ’=ln|π§| and π£=π+2ππ
βπ€=ln|π§|+πargπ§
This is the complex logarithm.
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ID: 190400003 Last Updated: 4/3/2019 Revision: 0
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