I've worked on solving this problem from my textbook for Non-Euclidean Geometry by following the process outlined in their example, however I got stuck on the simplification of the answer.

According to the formula, the equation is: $$\lvert\ln\frac{\bigl(1+2i-(2+\sqrt{5})\bigr)\bigl(i-(2-\sqrt{5})\bigr)}{\bigl(1+2i-(2-\sqrt{5})\bigr)\bigl(i-(2+\sqrt{5})\bigr)}\rvert$$

I got the answer, once simplified, to be: $$\lvert\ln\frac{-\sqrt{5}+5}{\sqrt{5}+5}\rvert$$

While my book has the simplified answer as: $$\ln\frac{\sqrt{5}+1}{\sqrt{5}-1}$$

After typing these all into WolframAlpha, they are all equal to the same decimal expansion, so I know that my answer is technically "correct", however I would greatly appreciate it if someone could explain how my textbook got to the simplified form, or how to get to the textbook answer from my current answer. Thank you so much!

Also in case this helps, WolframAlpha lists another form for all of these as: $$\ln\Bigl(\frac{1}{2}(3+\sqrt{5})\Bigr)$$

Question has been edited since I included a few typos.

  • $\begingroup$ They cannot be all equal: in your answer, the argument of the logarithm is $-1$. I presume that you omitted the logarithm in the book answer (please edit the question and fix that). Assuming that the logarithm should be present, that does indeed equal the last form that W-A gave you, but neither one of those equals your answer (unless you've mistyped it). $\endgroup$ – NickD Oct 22 '17 at 3:00
  • $\begingroup$ Thank you, I totally missed those typos. I meant to take off the abs on the books answer, instead of the abs and the log. The post has since been edited. $\endgroup$ – C Ordvia Oct 22 '17 at 4:29


Dividing the above expression by $\sqrt5$ to get your textbook answer.



I think you make some mistakes in your calculation. but the trick is there.

Your textbook just used the fact that

$$\sqrt a \times\sqrt a= a $$


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