Is it possible to use complex logarithm to integrate 1/(z+i) along a path? Evaluate the following on the path gamma_1 with endpoints [−1,1+i] I_1=(i)/(2)int_(gamma_1) (1)/(z+i)dz -(i)/(2)int_(gamma_1)(1)/(z-i}dz Am I allowed to use the antiderivative Log(z+i) and Log(z−i) respectively to evaluate I_1, or the am I restricted by the complex logarithm's analyticity problems.

Jamarcus Schroeder

Jamarcus Schroeder

Open question

2022-08-19

Is it possible to use complex logarithm to integrate 1 / ( z + i ) along a path?
Evaluate the following on the path γ 1 with endpoints [ 1 , 1 + i ]
I 1 = i 2 γ 1 1 z + i d z i 2 γ 1 1 z i d z
Am I allowed to use the antiderivative Log ( z + i ) and Log ( z i ) respectively to evaluate I 1 , or the am I restricted by the complex logarithm's analyticity problems.

Answer & Explanation

elgrupomentasb

elgrupomentasb

Beginner2022-08-20Added 9 answers

Yes, you are allowed to do what you want if you can find a branch of the logarithm that is holomorphic on a neighborhood of your curve. For example, if γ is a smooth curve from z 1 to z 2 along which log ( z + i ) is holomorphic, then d d z log ( z + i ) = 1 / ( z + i ), and
γ 1 z + i d z = 0 1 1 γ ( t ) + i γ ( t ) d t = 0 1 d d t log ( γ ( t ) + i ) d t = log ( γ ( t ) + i ) | t = 0 1 = log ( z 2 + i ) log ( z 1 + i ) .
Of course you cannot do this if the logarithm takes a jump on the path, because then the function log ( γ ( t ) + i ) is no longer continuously differentiable. But you're allowed to use a different brach of log for one of your integrals and another branch of the other. The justification is found in the Fundamental Theorem of Calculus for integrals of one parameter when you have an anti-derivative which is smooth on the curve.
kalkulusk2

kalkulusk2

Beginner2022-08-21Added 1 answers

You can show that I 1 = 0 using parametrizations γ 1 ( t ) and γ 2 ( t ) = γ 1 ( t ):
γ 1 1 z + i d z = 0 1 γ 1 ( t ) 1 γ 1 ( t ) + i d t
γ 2 1 z i d z = 0 1 γ 2 ( t ) 1 γ 2 ( t ) i d t = 0 1 γ 1 ( t ) 1 γ 1 ( t ) i d t = 0 1 γ 1 ( t ) 1 γ 1 ( t ) + i d t
But
γ 2 1 z i d z = γ 1 1 z i d z
So
γ 1 1 z + i d z γ 1 1 z i d z = 0

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