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`Complex FunctionβComplex Exponential FunctionβProperties`

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# Complex Function

## Complex Exponential Function

For the function, π(π§)=β―π₯cosπ¦+πβ―π₯sinπ¦, (where π§=π₯+ππ¦) is an entire (=analytic in β function.

Some of its properties:

• if π¦=0, then π(π§)=π(π₯+πβ0)=π(π₯)=β―π₯, so π agrees with the "regular" exponential function on β
• π(π§)=β―π₯(cosπ¦+πsinπ¦)=β―π₯β―ππ¦

By definition. The complex exponential function, β―π§, sometimes also denoted exp(π§), is defined by

`β―π§=β―π₯ββ―ππ¦, where π§=π₯+ππ¦`

## Properties

For the function, β―π§= β―π₯ββ―ππ¦, where π§=π₯+ππ¦:

• |β―π§|=|β―π₯||β―ππ¦|=β―π₯
• argβ―π§=arg(β―π₯β―ππ¦)=π¦(+2ππ, where πββ€)
• β―π§+2ππ=β―π₯β―π(π¦+2π)=β―π₯β―ππ¦=β―π§
• ```β―π§+π€=β―(π₯+ππ¦)+(π’+ππ£), where π§=π₯+ππ¦, π€=π’+ππ£  =β―(π₯+π’)+π(π¦+π£)=β―π₯β―π’β―ππ¦β―ππ¦  =(β―π₯β―ππ¦)(β―π’β―ππ¦)=β―π§β―π€```
• 1π§=β―βπ§, since β―π§β―βπ§=β―0=1
• β―π§ is an entire function.
• Derivative πβ²(π§):

Let π’(π₯,π¦)=β―π₯cosπ¦, π£(π₯,π¦)=β―π₯sinπ¦

Then ```π’π₯(π₯,π¦)=ππ₯cosπ¦;π£π₯(π₯,π¦)=ππ₯sinπ¦ π’π¦(π₯,π¦)=βππ₯sinπ¦;π£π¦(π₯,π¦)=ππ₯cosπ¦```

Thus πβ²(π§)=π’(π₯,π¦)+ππ£(π₯,π¦)=β―π₯cosπ¦+πβ―π₯sinπ¦=β―π§

So the derivative of β―π§ is β―π§, in symbols, `ddπ§β―π§=β―π§`.

• `ddπ§β―ππ§=πββ―ππ§ (πββ)` by the chain rule
• β―π§=β―π₯βππ¦=β―π₯β―βππ¦=β―π₯β―ππ¦=β―π₯β―ππ¦=β―π§
• β―π§=1 if and only if β―π₯β―ππ¦=1. The complex number in polar form, β―π₯β―ππ¦, equals 1, when its length equals 1 and its argument equals 0, ie.e. when β―π₯ and y=2ππ, πββ€. Thus

`β―π§=1βπ§=2πππ, πββ€`
• β―π§=β―π€ββ―π§βπ€=1βπ§βπ€=2πππβπ§=π€+2πππ

The function π€=β―π§ is a mapping from `β π§-plane ` to `β π€-plane `.

For the images of horizontal lines, πΏ={π₯+ππ¦0|π₯ββ} for fixed π¦0ββ. Then β―π§=β―π₯+ππ¦0=β―π₯β―ππ¦0, a line from origin but not equal with fixed angle.

For the images of vertical lines, πΏ={π₯0+ππ¦|π¦ββ} for fixed π₯0ββ. Then β―π§=β―π₯0+ππ¦=β―π₯0β―ππ¦, 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

• When β―π§=0
```β―π§=0ββ―π₯ββ―ππ¦=0 Note: β―ππ¦ has absolute value 1  ββ―π₯=0  βNever...!```
• For a given π§ββ\{0}, is there a π€ββ such that β―π€=π§? Writing π§=|π§|β―ππ and π€=π’+ππ£ this is equivalent to:
```β―π€=π§ββ―π’β―ππ£=|π§|β―ππ  ββ―π’=|π§| and β―ππ£=β―ππ  βπ’=ln|π§| and π£=π+2ππ  βπ€=ln|π§|+πargπ§```

This is the complex logarithm.

ID: 190400003 Last Updated: 3/4/2019 Revision: 0

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