# [Solved] Evaluation of a certain complex function

• I
Hi. I would like to ask regarding this function that keeps on cropping up on my study (see picture below).

What I did is simply substitute values for A and b and I noticed that it ALWAYS results to a real number. If possible, I would like to obtain the "non imaginary" function that is equivalent to this function.

Sadly, I have... no "advanced experience" with regard to dealing with complex functions. Can someone point me to some identity of sorts that would allow me to evaluate this function such that all the imaginary numbers will be canceled?

I tried using Euler's formula then I couldn't do much with it, except knowing that A is cos(x) and +/- 1 is sin(x) (yes, I know. Pathetic lol).

Thank you very much!

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Nevermind, managed to resolve the problem. I just had to take the ln(A+Bi) then do complex number division. Thanks!

S. G. Janssens
Nevermind, managed to resolve the problem. I just had to take the ln(A+Bi) then do complex number division. Thanks!

I don't know the details of your calculation and how important they are to you, but you may be interested in looking up the principal value of the complex logarithm, if you are not already familiar with it?

(Briefly, if ##c## is any real number and ##H(c) = \{z \in \mathbb{C}\,:\, c \le \Im{z} < c + 2\pi\}## then ##z \mapsto e^z## is bijective as a function from ##H(c)## to ##\mathbb{C} \setminus \{0\}##. So, for every choice of ##c## there exists an inverse which we might denote by ##\ln##. However, usually this is reserved for the particular inverse corresponding to the choice ##c = -\pi## and then ##\ln## is called the principal value of the logarithm.)

If you use the principal value logarithm, then you find indeed that ##\ln{\frac{A + i}{A - i}} = 2 i \arg{(A + i)} \pmod{2\pi i}## with ##\arg(A + I) \in [-\pi, \pi)## so if ##A## and ##b## are real, then so is your expression.

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