Complex Logarithm

Hello guys! Last time we left with the topic, logarithms where we said the logarithms are not generally used for negative base and argument as they can create some hindrance in indicating it on a 2 dimensional graph. On the other hand, we saw that graph of the log values of a positive base and positive argument shows a continuous curve line. But today, we will talk about the log value of  what is generally considered unrealistic. 

So far we have learn :

     log_ba = c  i.e. b^c = a

(log function is actually inverse of exponent function)

 

A single branch of the complex logarithm. The hue of the color is used to show the arg (polar coordinate angle) of the complex logarithm. The saturation and value (intensity and brightness) of the color is used to show the modulus of the complex logarithm.      

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Now, if you have an exponent function,

                          -2¹ = -2 

                       i.e.  log_-2 -2 = 1

                  or ln -2/ln -2 = 1

                                                                    

 using natural log , 

       ln -2 =?   Or       e^? = -2 

       since, e^iπ +1 = 0

               e^iπ= -1

     e^iπ + e^iπ = -1 + -1 = -2

         ln( e^iπ + e^iπ )

         =  ln(2.e^iπ) 

         = ln2 + ln e^iπ 

    = ln2 + iπ                            

 ( log_e e^iπ = ln_eiπ = iπ , since : log_b b¹ = 1)

  = ln2 + iπ (2m+1)                

 ( the angle can be obtained periodically i.e. 90° is equivalent to 360° + 90°  and  m.360° + 90°  )

        Here you can simply substitute the value for fetching the answer quickly 

 ln(-a) = ln(a) + ln(e^iπ)       

  or       ln(-a) = ln(a) + iπ

      Where a>0

Let’s take another example,

  -2³ = -8    i.e.     log_-2-8 = 3

Or   ln-8/ln -2 =             

    ln(-1. 8) / ln(-1 . 2)

=  ( ln8 + ln -1)  / ( ln2 + ln-1 )

                             

 = ( ln 2³ + ln-1) /( ln2 + ln-1)

                              

 = (3 ln 2 + ln -1)/ ( ln2 + ln-1)

 = (3 ln 2 + iπ (2m +1) )/ ( ln2 + iπ (2n+ 1))                since e^iπ= -1             WHERE  m, n € Z but m is not equal to n

if m =1 and n= 0

( 3 ln 2 + iπ (2 . 1 + 1)  ) /( ln 2 + iπ (2 . 0 + 1 ) ) = ( 3 ln 2 + iπ (2 + 1)  ) / ( ln 2 + iπ ( 0 + 1 )

= ( 3 ( ln 2 + iπ ) ) /  ( ln 2 + iπ ))

    =                         3   

So 3  is one of the answers when we put m = 1 and n = 0

Here it is clear that log_-2 -8 is a multi  valued function

So on the basis of this now we can even find the value of log₂ -8 

                     ln-8/ln 2 =

                          ln(-1. 8) / ln( 2)

       =    ( ln8 + ln -1)  / ( ln2)

      =     ( ln 2³ + ln-1) /( ln2)

      =     (3 ln 2 + ln -1)/ ( ln2)

     =     (3 ln 2 + iπ (2m +1) )/ ( ln2)                since e^iπ= -1                                                                                                                             

    3 + iπ ( 2m + 1)/ ln2

And this can bring many answers to this function

Again if  we have to find ln(z), where 

                                    z = x + i y :

                                                                                         

       ln (z) =1/2  .  ln( a² + b² )  + i tan^-1 (b/a) 

              if you have studied phasor algebra, you can simply understand that √(a² + b²⁾is a modulus and coefficient of i is an angle.

So you can simply find the log of any given value , whether it is a complex value or real value or imaginary value.

As I have mentioned the method of finding the log of a complex (the value ( x + i y ) is somewhat a rectangular form ) value above, which is similar to a Cartesian  coordinate if a comma ‘ ,’ is used between X and Y 

Other than that there are 3 more forms mentioned below :

                                                                                 2 - triangular form: r ( cosΦ + i sinΦ)

                                                                                   3 – polar  form :   ( r ∠Φ )

                                                                                  4 -  exponential form:    ( r e^iΦ )

So if any of the form is given find the log of the modulus and find the angle to be written as a coefficient of imaginary part i. otherwise find the log of a real number if Imaginary part is not present. That's it, Ciao!