Confirming the Euler's Formula: e^{i(a+bx)}=

[dave4000]

Is this true:

$$e^{i(a+bx)}=cos(a+bx)+i sin(a+bx)$$ ?

 

[chroot]

Yes -- that's the beauty of function notation. You can replace 'x' with any expression you like.

- Warren

 

[dave4000]

Thanks

 

[Scandal]

I know Daves question is answered, but instead of making a new thread I just ask another one here as it is somewhat related. Unfortunately I don't have a mathtype program so you'll have to imagine.
So if you have the e^xi expression and set x = pi, this will equal -1. Adding 2Pi equals 1. So far so good? But here comes what I don't understand. 1 can also be written e^0, so using logarithms we say: e^i2Pi = e^0 -> i2Pi = 0, but this can't be true as i2Pi = 6,28i. I got one possible solution from my math teacher but it did not quite tell me how it could equal 0. Anyone care to give an explenation a shot?

 

[AUMathTutor]

My best guess would be that the exponential function is not a bijection over the complex numbers, while it is so over the real numbers. That is, for real numbers,

e^x = e^y <=> x = y is true.

For complex numbers, it must be the case that

e^x = e^y <=> x = y is false.

This shouldn't be so surprising. For the real numbers, you can't say things like...

x^2 = y^2 <=> x = y
sin(x) = sin(y) <=> x = y
...

I suppose it's just an accident that you can't do that for the exponential function over complex numbers.

 

[dx]

My best guess would be that the exponential function is not a bijection over the complex numbers, while it is so over the real numbers. That is, for real numbers,

e^x = e^y <=> x = y is true.

You mean injection.

 

[AUMathTutor]

Well, injection may be true too, but it's definitely not a bijection either.

I tend to talk in terms of bijection or not bijection. I usually don't delve into things like injectivity and surjectivity. I think that the exponential is neither bijective nor injective over the complex numbers... then again, what do I know?

 

[dx]

You said exp was a bijection over the reals which it isn't. It's an injection, but not a bijection.

 

[Count Iblis]

The problem is not with the exponential function, but with the logarithmic function. You need to define a branch cut to define the logarithmic function. This then means that the imaginary part of the Log of a complex number (the so called argument) is a unique number in some interval of length 2 pi.

 

[belliott4488]

Yeah, what Count Ibis said:

The problem is not with the exponential function, but with the logarithmic function. You need to define a branch cut to define the logarithmic function. This then means that the imaginary part of the Log of a complex number (the so called argument) is a unique number in some interval of length 2 pi.

For the less mathematically versed, this is a little like taking the square root of a number and having to remember that there are actually two roots (+/-). When you take the log of a complex number, there are (potentially) multiple correct results, so you have to specify which one you're choosing.

(At least, I think that's a correct statement; someone please correct me if it's not.)

 

[AUMathTutor]

"You said exp was a bijection over the reals which it isn't. It's an injection, but not a bijection."

Oops. You're right. I guess that's the price you pay when you get too used to using the general terminology too often.

 

[Tibarn]

The trouble is that the complex exponential function is periodic, so saying that e^x = e^y iff x=y for complex x and y is akin to saying that cos x = cos y iff x=y.