hmmmmm: explain e^(pi*i) = e^(-pi*i) = -1 ==> (e^(pi*i))^i = (e^(-pi*i))^i ==>...
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pka
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For me it is hard to know exactly what help you need?
The part in red is FALSE. This is the classic example that shows
the some of the usual(real no.) rules do not work for complex numbers.
Is it the case that you do not understand the definition of complex exponents?
\(\displaystyle \Large z^w=\exp \left( w \log(z) \right)\) that is the principal value.
Recall that \(\displaystyle \Large \log(z)=\log(|z|)+{\bf{i}}\cdot\text{Arg}(z)\)
If you have other more basic issues, then you need a sitdown with a tutor. That is not a function of this site.
D
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Those (laws of exponents) also do not apply for 1, -1 and 0.
(-1)2n = (1)2n does NOT imply 1= (-1)
(-1)2n = (1)2n does NOT imply 1= (-1)
Correct. However, if a = b, then in R, ar = br. And therein lay my confusion.
In the case at hand, a = epi*i = -1, and b = e-pi*i = -1. Therefore, a = b and i = i, but it doesn't necessarily follow that ai = bi. Hence it appears that exponentiation is not well defined (in the usual sense) in the complex plane. Certainly I was not suggesting that epi is equal to its reciprocal. I just found the result counterintuitive and thus sought explanation from a professional. Enjoy the night.
pka
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We are not to teach you the basics. You post what you do not understand along with what you have done and we try to sort it out.
But if you ask a question about complex exponents but know nothing about complex exponents then that is an example of asking to be taught not just helped. The function of the site is to help not to teach basics. That is what it means.
Do you have any idea how to evaluate \(\displaystyle (-1)^{\bf{i}}~?\)
D
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These (complex) are multi-valued function - these will fail vertical-line test.
ei*π = ei*2*π = ei*22*π does NOT mean 1 = 2 = 22
Do you have any idea how to evaluate [FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main-bold]i[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main]?[/FONT]
(-1)i = (epi*i)i = epi*(i^2) = e-pi
I am of retirement age and relearning (on my own) the mathematics I learned long ago. I had simply hoped someone could explain what I was seeing as a contradiction. I shall not ask again.
(-1)i = (epi*i)i = epi*(i^2) = e-pi
I am of retirement age and relearning (on my own) the mathematics I learned long ago. I had simply hoped someone could explain what I was seeing as a contradiction. I shall not ask again.
pka
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Why did you fail to provide us with all the background information in the first place.
We agree that \(\displaystyle e^{i\pi}=(-1)=e^{-i\pi}\).
If we could use the real laws of exponents you should think that \(\displaystyle \left(e^{i\pi}\right)^i=e^{i^2\pi}\) must be true. BUT it is not true. If it were true then \(\displaystyle e^{\pi}=e^{-\pi}\) which is clearly false. So there is a counter-example, some of laws that work for real numbers do not work for complex numbers.
That is what we both wanted you to see. Moreover, we assumed that you were a student in a class learning complex number operations.
PKA and Subhotosh Khan, thank you both for the clarification.
PKA, upon rereading my initial post, I must agree that my specific problem was poorly stated.
The real issue was in choosing the correct "principle argument". That said, let's see whether I have this right once and for all.
z = a + bi = |z| [cos(t) + i sin(t)] = |Z| eit, where argument t is in [-pi/2, 3pi/2). That said, in evaluating (-1)i, we have (epi*i)i. That is, the principle argument is pi (not -pi) ... hence the apparent ambiguity of my initial post is resolved. Please tell me I have this right.
As another example, sqrt(-i) = [e-pi/2 * i ]1/2 = e-pi/4 * i = 0.5sqrt(2) [1 - i]. By contrast, had I chosen argument 3pi/2, I would obtain the non principal square root -0.5sqrt(2) [1 - i]. How's my thinking on this?
Again, thank you so much.
Rich
PKA, upon rereading my initial post, I must agree that my specific problem was poorly stated.
The real issue was in choosing the correct "principle argument". That said, let's see whether I have this right once and for all.
z = a + bi = |z| [cos(t) + i sin(t)] = |Z| eit, where argument t is in [-pi/2, 3pi/2). That said, in evaluating (-1)i, we have (epi*i)i. That is, the principle argument is pi (not -pi) ... hence the apparent ambiguity of my initial post is resolved. Please tell me I have this right.
As another example, sqrt(-i) = [e-pi/2 * i ]1/2 = e-pi/4 * i = 0.5sqrt(2) [1 - i]. By contrast, had I chosen argument 3pi/2, I would obtain the non principal square root -0.5sqrt(2) [1 - i]. How's my thinking on this?
Again, thank you so much.
Rich
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Just to add my 2 cents: This kind of behavior is (partially) covered when learning about Riemann sheets, branch cuts, and principle values. Thus, if the branch cut is along the negative x axis [\(\displaystyle -\pi\, \lt\, arg(z)\, \le\, \pi\)], there is a discontinuity there, i.e. arg(-1) \(\displaystyle \to\, \pi\) if one approaches -1 from on or above the branch cut which is on the main (principal) sheet. And arg(-1) \(\displaystyle \to\, -\pi\) if one approaches -1 from below the branch cut on the main sheet which is not on the main sheet. So, in that sense
-1 = \(\displaystyle e^{i\, \pi}\) on the main Riemann sheet is not the same as
-1 = \(\displaystyle e^{-i\, \pi}\) on a different Riemann sheet.
EDIT: Oh, BTW we actually have
\(\displaystyle -1\, =\, e^{(2n+1)\, i\, \pi}\)
so that
\(\displaystyle (-1)^i\, =\, (e^{(2n+1)\, i\, \pi})^i\, =\, (e^{-(2n+1)\, \pi})\)
depending on just which sheet you choose -1 to be [which value of the integer n you choose].
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pka
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I now understand you background and purpose. So what I have to say may appear to be very esoteric.
About exactly thirty years ago in the mid 1980's came an effort at reform in mathematics education.
In complex variables, growing out of model theory, it was thought that by defining \(\displaystyle \bf{i}\) as a number that solved the real equation \(\displaystyle x^2+1=0 \) we enlarged the number system to the complex case. By doing so we eliminate any talk of \(\displaystyle \bf{i}=\sqrt{-1} \) (that was now OUT!) In fact, \(\displaystyle \sqrt x \) is only defined for real numbers greater than or equal zero.
You asked about \(\displaystyle \sqrt{\bf{i}}\). But we don't use that notation.
It is true that there are two square roots of \(\displaystyle \bf{i} \), \(\displaystyle \exp\left(\dfrac{\bf{i}\pi}{4}+k\pi\right),~k=0,1 \).
Next you said something about principle argument, that notation is \(\displaystyle -\pi<Arg(z)\le\pi \). note the capital A
Suppose that \(\displaystyle x\cdot y\ne 0 \) then
\(\displaystyle \text{Arg}(x + yi) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right. \)
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