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April 25th, 2016, 05:58 PM  #1 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0  Why is this true?
So, I just started typing random stuff into my Ti84 and found that this seems to be true for all but x=0 (since it's undefined): = = Re[x^i] (Where Re[a+bi] is a function that gives only the real part (a) of any complex output.) (The "log" refers to the natural log, base e, NOT base 10.) Can somebody please explain to me why this is? I really have no idea. I'm also not sure why 0^i is undefined either. Last edited by John Travolski; April 25th, 2016 at 06:04 PM. 
April 25th, 2016, 06:56 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,509 Thanks: 2514 Math Focus: Mainly analysis and algebra  One definition says that $\newcommand{e}{\mathrm e}a^b = \e^{\log a^b} = \e^{b \log a}$, but this is only valid for $a \gt 0$ in the real numbers, because $\log a$ is defined only for $a \gt 0$. However, we can assign a value to $0^b$ for positive rational numbers $b$, by using (and extending) the definition that $a^b = \underbrace{a \times a \times \ldots \times a}_{b \, \text{terms}}$ where $b$ is a natural number. And by the use of the limit definition of real numbers, we then arrive at $0^b=0$ for all positive real numbers $b$. In fact we have $\lim \limits_{a \to 0} a^b = 0$ and $\lim \limits_{x \to b} a^x = 0$ both for all positive $b$. We might think that this idea might be of use in the complex plane too. But when we write $b=u+iv$ we get $a^b= \e^{(u+iv)\log a} = \e^{u\log a}\e^{iv \log a}$ Now, if $u \ne 0$ here, we get $\lim \limits_{a \to 0} \left\e^{u\log a}\e^{iv \log a}\right = \left\lim \limits_{a \to 0} \e^{u\log a}\right \left\lim \limits_{a \to 0} \e^{iv \log a}\right = \left\lim \limits_{a \to 0} \e^{u\log a}\right = 0$ and so $\lim \limits_{a \to 0} \e^{u\log a}\e^{iv \log a} = 0$. But if $u=0$ we have $\lim \limits_{a \to 0} \e^{u\log a}\e^{iv \log a} = \lim \limits_{a \to 0} \e^0 \e^{iv \log a}=\lim \limits_{a \to 0} \e^{iv \log a} $. And as $a \to 0$ we have $\log a \to \infty$ and so $\e^{iv \log a} $ just rotates around the unit circle in the complex plane and doesn't converge. So $0^{iv}$ is undefined for any $v$. 
April 26th, 2016, 01:29 PM  #3 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 
I appreciate the explanation for why 0^i is undefined. However, does anybody know why cosine shows up in what I mentioned before that?

April 26th, 2016, 04:51 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,958 Thanks: 1845 
The cosine shows up because of Euler's formula.

April 26th, 2016, 06:06 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,509 Thanks: 2514 Math Focus: Mainly analysis and algebra 
$\newcommand{e}{\mathrm e} x^i = \e^{\log x^i} = \e^{i \log x} = \cos \log x + i \sin \log x$ where $x \gt 0$. If $x < 0$ then $x = r\e^{i\pi}$ where $r=x$ and so $x^i = r^i \e^{\pi} = \e^{\pi}\left(\cos \log r + i \sin \log r \right) = \e^{\pi}\left(\cos \log x + i \sin \log x \right)$. Last edited by v8archie; April 26th, 2016 at 06:13 PM. 

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