R Thread Starter rahulpsharma Joined Sep 5, 2010 54 Nov 29, 2020 #1 I am unable to spot the error in this calculation...!! What's wrong here...?? Thanks and regards, Rahul
I am unable to spot the error in this calculation...!! What's wrong here...?? Thanks and regards, Rahul
WBahn Joined Mar 31, 2012 26,432 Nov 29, 2020 #4 Wolframore said: The mistake is here View attachment 223647 Click to expand... So what is the mistake? Is it that (a^x)^2 isn't a^(2x)? Or is it that (-1)^2 isn't 1? If you are being careful, it should be noted that \(e^{ix} \; = \; e^{i\left(x+2n\pi\right)}\) I would say that the problem is here: \(ln \left( e^{2 \pi i} \right) \; = \; 2 \pi i\) This should be \(ln \left( e^{2 \pi i} \right) \; = ln \left( e^{2n \pi i} \right) \; = \; 2n \pi i \;\; \text{(n is an integer)}\) So now in the next set you have \(2n \pi i \; = \; 0 \) which is true if we choose n=0
Wolframore said: The mistake is here View attachment 223647 Click to expand... So what is the mistake? Is it that (a^x)^2 isn't a^(2x)? Or is it that (-1)^2 isn't 1? If you are being careful, it should be noted that \(e^{ix} \; = \; e^{i\left(x+2n\pi\right)}\) I would say that the problem is here: \(ln \left( e^{2 \pi i} \right) \; = \; 2 \pi i\) This should be \(ln \left( e^{2 \pi i} \right) \; = ln \left( e^{2n \pi i} \right) \; = \; 2n \pi i \;\; \text{(n is an integer)}\) So now in the next set you have \(2n \pi i \; = \; 0 \) which is true if we choose n=0
Wolframore Joined Jan 21, 2019 2,549 Nov 29, 2020 #5 solve for (e^(i pi))^2 the issue is you're using complex "i"