Is 5pi/2 equal to pi/2 modulo 2pi?

[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.

 

[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.

 

[Landau]

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

 

[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??

 

[laohu]

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

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

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.

 

[Mute]

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

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.

 

[ohubrismine]

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.

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.

 

[Gregg]

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.

Is this a serious question?

 

[The Chaz]

Is this a serious question?

Is this a serious answer?

 

[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

 

[lm_aquarius]

Be careful about the domain of the functions.

 

[G037H3]

is 5x/2=x/2 ?

 

[Mentallic]

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

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.

 

[AC130Nav]

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.

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.

 

[Mentallic]

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.

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.

 

[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.