# Is 5pi/2=pi/2 ?

1. Jun 15, 2010

### coki2000

Hi,

$$i=e^{\frac{\pi }{2}i}=e^{\frac{5\pi }{2}i}\Rightarrow e^{\frac{\pi }{2}ii}=e^{\frac{5\pi }{2}ii}=e^{-\frac{\pi }{2}}=e^{-\frac{5\pi }{2}}$$

The result of this is 5pi/2=pi/2.Please explain to me.

2. Jun 15, 2010

### eok20

You need to be careful with the function a -> a^i since you need logarithm to define it and therefore need to stick with one branch.

3. Jun 16, 2010

### Landau

exp(a)=exp(b) does not imply a=b, but a=b+i2pi.

4. Jun 16, 2010

### The Chaz

cos(x)=cos(-x).
That is a true statement for all real values of x.
So x = -x for any real value of x, which implies that every real number is actually EQUAL TO ZERO!!!

Do you (OP) understand the error in this??

5. Jun 16, 2010

### laohu

The problem here is that the exponential function on the complex plane is not injective. That is, even if ea = eb, we cannot infer that a = b. Analogously, even if sin(0) = sin($$\pi$$), we cannot infer that 0 = $$\pi$$.

Probably a typo there. a and b differ by a multiple of 2 i pi, i.e. a = b + 2k i pi where k is an integer.

Last edited: Jun 16, 2010
6. Jun 17, 2010

### Mute

This is not quite a resolution to the OP's question (even fixed to read $2\pi k$ instead of just $2\pi$). If a and b are real, that still implies a = b, and k = 0.

The fully correct resolution is, as eok20 stated, that stating $i = e^{something}$, what you're really doing is writing $i = \exp[\log i]$, so you need to choose a branch of the logarithm. In doing so, you cannot equate $\exp(i\pi/2)$ and $\exp(5\pii/2)$ because they exist on different branches.

7. Jun 22, 2010

### ohubrismine

The expression of a complex number in exponential form is based on Euler's formula relating it to its polar form: r*exp(i*a)=r*[cos(a)+i*sin(a)]. Because the trig functions are periodic, there is not a unique polar representation. As already stated, there is no injective map.
Accordingly, the polar form of a complex number is defined based on the principal value of the argument of the complex number, which means restricting a to the interval (-pi,pi].
So, fundamentally, it is not correct to say that i= exp(i*pi/2) = exp(i*5pi/2).
However, it so happens that 5pi/2 = pi/2 since 5pi/2= pi/2 + 2pi on the unit circle.

8. Jun 22, 2010

### Gregg

Is this a serious question?

9. Jun 22, 2010

### The Chaz

Is this a serious answer????

10. Oct 20, 2010

### GeoFiend

Hi, I am currently doing Logarithmic Differntiation in my class and had to simplify a similar answer and was able to do so by takeing the natural log of both sides.
ex: e^x=e^pi/2
lne^x=lne^pi/2
x=pi/2

11. Oct 20, 2010

### lm_aquarius

Be careful about the domain of the functions.

12. Oct 20, 2010

### G037H3

is 5x/2=x/2 ???

13. Oct 21, 2010

### Mentallic

Your point being? Obviously $x=\pi/2$ since the exponents need to be the same. This thread is talking about exponentials in the complex plane, which is a little more complicated.

14. Oct 24, 2010

### AC130Nav

Or maybe the question is simple, and the student is just confused by the results. 5pi/2=pi/2 because 5pi=pi because pi is 180 degrees and all the multiples of 2 pi drop out.

Just a thought.

15. Oct 24, 2010

### Mentallic

No $$5\pi/2\neq \pi/2$$, obviously. What you're referring to is trigonometry such that $$sin(\pi/2)=sin(5\pi/2)$$ and such. This is not the same.

I'm still confused as to what GeoFiend was trying to accomplish with his post.

16. Oct 25, 2010

### HallsofIvy

Just to stick in my oar: $\pi/2= 5\pi/2$ modulo $2\pi$. And, if you are working with trig functions that have period $2\pi$, then "modulo $2\pi$" is sufficient.

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