My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
April 25th, 2016, 05:58 PM   #1
Senior Member
 
Joined: Oct 2015
From: Antarctica

Posts: 128
Thanks: 0

Lightbulb Why is this true?

So, I just started typing random stuff into my Ti-84 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.
John Travolski is offline  
 
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
Quote:
Originally Posted by John Travolski View Post
I'm also not sure why 0^i is undefined either.
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 $v$.
v8archie is offline  
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?
John Travolski is offline  
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.
skipjack is offline  
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.
v8archie is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
true



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
X^n+Y^n=Z^n is true but not in N M_B_S Number Theory 16 December 16th, 2014 12:37 AM
It is true that masterofmath Algebra 1 August 31st, 2014 10:19 AM
True or not Dacu Number Theory 16 June 19th, 2013 06:42 AM
Is this true ????? johnmath Real Analysis 1 July 5th, 2011 08:53 AM
Is this true? sivela Calculus 1 February 15th, 2010 12:41 PM





Copyright © 2018 My Math Forum. All rights reserved.