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Pr0bz

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e^(i*pi)+1

(i är den imaginära enheten, inte en variabel)

Hjälp mig lösa det =)
 
Pr0bz skrev:
e^(i*pi)+1

(i är den imaginära enheten, inte en variabel)

Hjälp mig lösa det =)
Se nedan.
e^(i*pi) = -1: pi = 0 ?

Date: 10/17/97 at 11:29:15
From: John K. Koehler
Subject: Weird tricks

Dr. Math,

I know that from a certain trig identity we get the equation
e^(i*pi) = -1. I have been playing around with this equation and have
found some disturbing things. I hope that you can help me.

First:

e^(2*i*pi) = 1
e^(-2*pi) = 1 (raised both sides to i and 1^n is 1)
-2*pi = 0 (took the ln of both sides and ln(1) = 0)
pi = 0 !

Second:

i*pi = ln(-1) {1}
ln(-n) = ln(-1) + ln(n)
therefore by {1}
ln(-n) = i*pi + ln(n)

Finally:

i^2 = -1
e^(i*pi) = i^2
e^(-pi) = i^(2i)
e^(-pi/2) = i^i

e^(-pi/2) is a real number and therefore so is i^i

Can you please tell me why I get such ridiculus results?
I asked my calc2 teacher last year - but all he could tell me
was that imaginary numbers don't work like reals, and I still
don't see how that would change anything.

Thanks in advance,
John
-----------------------------------------------
Date: 10/17/97 at 13:08:36
From: Doctor Bombelli
Subject: Re: Weird tricks

Well, John, not all the results you have are ridiculous! In fact, you
and your teacher are both right; complex numbers don't always work
like real numbers, but sometimes they do.

Complex numbers can be written in many equivalent ways:

z = x+iy
z = re^*(it) where t is the angle the point (x,y) makes with the
positive x axis.
t is called the argument of z--arg(z)
z = r[cos(t)+i sin(t)]

In fact, the definition of e^(it) is cos(t)+i sin(t).

This is why we have e^(-ipi) = cos(pi)+i sin(pi)= -1.

We would like w = log(z) and z = e^w to match up, just as for real
numbers.

Let z = re^(it) and w = u+iv.

z = e^w = e^(u+iv) = e^u*e^(iv) and z is also re(^it)

Match the real part and the "i part" (the imaginary part).

r = e^u and e^(it) = e^(iv)

So u = ln(r) and v can be any value with cos(v) = cos(t)
[since e^(it) is cos(t)+i sin(t)].

So we say, for complex numbers, ln(z)=ln(r)+i arg(z)+i2kpi
(k an integer). In fact, ln(z) has many values.

We also define, in this way, that

z^a = e^[a ln(z)] when a is complex.

(Check that this works for real numbers a, also.)

Note that 1^a = e^[ln(1)] = e^[0+i2kpi], so only one of the answers
is e^0 = 1.

Here is your problem:

e^(2*i*pi) = 1
e^(-2*pi) = 1 (raised both sides to i and 1^n is 1)
-2*pi = 0 (took the ln of both sides and ln(1) = 0)
pi = 0

In steps 2 and 3 you don't have one-to-one functions any more.
(The reason e^x = 1 implies x = 0 is because the real logarithm
function is one-to-one. It is kind of like why x^3 = 1 gives x = 1
and x^2 = 1 gives x = 1 or -1. The cube root is one-to-one, but
the square root isn't.)

Now in your second experiment:

i*pi = ln(-1) {1}
ln(-n) = ln(-1) + ln(n)
therefore by {1}
ln(-n) = i*pi + ln(n)

This is okay.
ln(-n) = ln(n) + i pi + i 2kpi from the definition above.

In your third experiment:

i^2 = -1
e^(i*pi) = i^2
e^(-pi) = i^(2i)
e^(-pi/2) = i^i

e^(-pi/2) is a real number and therefore so is i^i

i^i is e^[i ln(i)], and ln(i) is ln(1) + i pi/2 + i 2kpi

So... i^i = e^[0 + i^2 pi/2+ i^2 2kpi] = e^[- pi/2 - 2kpi]
which is a real number.

I commend you on playing around with this stuff. That is how new
mathematics is learned, on all levels.

-Doctor Bombelli, The Math Forum
 
Vad blir svaret då? Enligt deras uträkning blir det väl 1 på min fråga? :S
 
Så genuint ointelligent jag känner mig helt plötsligt..... Som ett ton tegel eller nåt.
 
Pr0bz skrev:
e^(i*pi)+1

(i är den imaginära enheten, inte en variabel)

Hjälp mig lösa det =)

Om det är så som du skrivit e^(i*pi)+1 och inte e^((i*pi)+1)


r*e^(i*A)=r(cosA+i*sinA)
r=1
A=pi
alltså
e^(i*pi)+1 = (cos(pi)+i*sin(pi))+1 = 0 + i*1 + 1= 1+i
 
Last edited:
Ooops!
Kom på en sak förut. Jag räknade fel inatt (händer mycket sällan).
Här är den riktiga lösningen:
e^(i*pi)+1 = (cos(pi)+i*sin(pi))+1 = -1 + i*0 + 1= 0
Sorry.
 
jo då så att.........jag har en tjejkompis vars hund gillar matte!


(Ber om ursäkt för det meningslösa inlägget, men jag brukar skojja bort saker jag inte förstår! Mina vänner säger att jag är väldigt skojjig!)
 
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