# 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.

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 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

 Why did I what? What assumption? – alvoutila Aug 27 '12 at 12:24 @alvoutila: You included a link to WolframAlpha, claiming that it "shows they are the same". It shows nothing of the kind; it shows that WolframAlpha returns a certain antiderivative for a certain indefinite integral, but you mentioned none of that or any of the steps that you took in incorrectly concluding that this is equivalent to showing that these two expressions are the same. That's very misleading, and unnecessarily so; you could have described directly what the WolframAlpha link shows, and what you concluded from that and why. – joriki Aug 27 '12 at 12:49 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 – alvoutila Aug 27 '12 at 16:18