# Is $(-1+i)\log(2e^{it}+i)$ same as $\frac{1}{2}\left((2+2i)\tan^{-1}(2e^{it})-(1-i)\log(1+4e^{2it})\right)$?

Is $\displaystyle(-1+i)\log(2e^{it}+i)$ the same as $\displaystyle\frac{1}{2}\left((2+2i)\;\tan^{-1}(2e^{it})-(1-i)\log(1+4e^{2it})\right)$?
WolframAlpha shows that they are same, but this page on W|A shows FALSE so they are not same.

-
Advice: use www.tinyurl.com , or any other site of the kind, to shorten such large addresses, otherwise it gets messed up. –  DonAntonio Aug 25 '12 at 15:49

well this link [WolframAlpha][1] shows that $(-1+i)\log(2e^{it}+i)$ and $\frac{1}{2}(2+2i)\tan^{-1}(2e^{it})-(1-i)\log(1+4e^{2it})$"... is equivalent for restricted values of $t$..." [1]: wolframalpha.com/input/?i=integrate%28%20%28-2*e%5E%28i*t%29-2*i*e%5E‌​%28i*t%29%29/%282*e%5E%28i*t%29%2bi%29,%20t%29 –  laovultai Aug 27 '12 at 16:18