@janhardo 

Ok, i posted a double post ..it slipped in and the post of @nm 6131 and @kitonum are removed

The question was there was how to rearrange this  complex ln expression 

As you can see i did it by hand too, but don't know if it is right? :

If i remember the answer from nm 6131 it was like this :
expr : = I*ln(-z*I + b) - ln(z*I + b); 
I*simplify(log(exp(1)*(expr)) ;   ?        (in general : ln x^a = a*ln x)  or  take ln^a             

Note: should be nice  to  have the answer of @nm 6131 and @kitonum back again

@Kitonum 

Thanks

Your calculation brings  me back to the direction of the start formula 

@nm 

Thanks

Seems that mathematica is more advanced then Maple? 
Well, i stay  with Maple, because learning again a new language for Mathematica is too much. 

Surprize, the manual derivation of the ln expression seems to be correct what i did by hand. (it is the same as your Maple calculation)
But your  code example shows me that i must practice with this ..

Not capable of not using advanced commands makes it neccesary to know the basic rules 
Its here : a power of a logarithm with base e : 

e 
  ln x^n =  n.ln x 

The answer of @acer is almost the same as in the picture 
Now i tried it doing also, but with easier commands... how far do i come to the wanted answer? 

Download handmatig_betrekking_logaritmisch_en_goniometrisch_forum_nm_6131.mw

@Carl Love 

Thanks

It is clear that after reading your step by step explanation i ever never could deduct this by meself, the working of combination of commands.
For instance could not give meaning to [Re,Im](J1) and the nested simplify command.
Very helpful to try understand the whole picture better with all details involved.

@Carl Love Thanks

It seems that the base-10 logarithm is not the logarithm be found in higher level mathematics..
A ln logarithm use is widespread,why is that, i can't give the answer .
The number e is popping up everwhere in math.

The derative value of e^x is e^x value for x
Also the e^x function can never be 0 
Its a remarkable  number e

@acer  Thanks

Download handmatig_betrekking_logaritmisch_en_goniometrisch.mw

@acer

Thanks

So your claim about my Maple result being "almost correct" except that I "is still there" is not right. 

---------------------------

I overlooked the  i  ..it is  in there  , you are right. ( comes that i am so used at the normal variables naming) 
I think you have checked that your answer with simplify if it was correct ?
Cannot recognize yet the book answer with your answer unfortanely if both are the same 

@vv 

Informative website for me getting more insight in the complex analysis

Complex Analysis (complex-analysis.com)

@acer

This answer you got has  minus sign and a I inside , but  the bookanswer has no  - and a I next to  ln  
It proves how you must be careful to type the right expression, but is all a little bit complicated to get this expression not in Maple input here as post 

@janhardo 

Download complex_integration-tom_apostel_example_-bernouille.mw

@acer 

Thanks

Your code is also a difficult one to understand quick , so perhaps you can explain this in your own words too for me to get a idea for me.

It was tricky, because this  expression written (see hereunder)i s not the same in Holland or America/England? 
Log and ln are the same in America ( and in the book example too probably) and not in Holland
Log and ln are different. 

So log written in the formula must read as ln then.
Your answer is almost correct , its only the I (imaginairy unit) is still there

In a another example of calculating a complex integral is used to remove the I   ....see hereunder

J:= Int(exp(t*x), x);
J1:= subs((SUB:= t= alpha+beta*I), J);
J2:= evalc([Re,Im](J1));

@Carl Love 

Thanks

There will be a reason why maple uses  log and ln for the same thing ..yes use in English math books ? 
In written math books here in Holland there is a distinction between the two: log and ln and they differ in meaning
Probably not in English math books : log and ln are the same then.

@Carl Love 

Thanks

I noticed the difference between input and output coloring, so its a mistake 
This second code piece construct the two wanted integrals from 

 J := Int(exp(t*x), x)

Amazing second code piece ,but it needs some clarifcation compared with the first code piece. 

@Carl Love 

I used 2d input expression palette and was writing , so it is not executable math then.
Was not further thinking on it , but it is possible to use the log and  ln for the same logaritme with  base e  : 

log[10](100);    can be used also in worksheet .
                             2
The book example use log as notation ( i assume with base 10 ) 

A  2 d input example                         

log[10](100);    
               2
ln(100.0);

             4.605170186
 

@janhardo 

Complex integration is somewhere started and ...

-------------------------------------------------------------------------------------------------info------------------------

So far, we’ve seen how to evaluate integrals of simple functions of a complex variable—that were defi ned in terms of a single real parameter we called t. Now it’s time to generalize and consider a more general case, where we just say we’re integrating a function of a complex variable f (z ), where ∈ C. This can be done using a technique called contour integration. The reason integrals of complex functions are done the way they are is that while an integral of a real-valued function is defi ned on an interval of the line, an integral  of a complex-valued function is defi ned on a curve in the complex plane.