Understanding the Solutions to e^x=1: Seeking Help

  • Thread starter Thread starter estro
  • Start date Start date
estro
Messages
239
Reaction score
0
This is not a homework question.

I know that the equation e^x=1 has infinitly many solutions in form of 2i\pi n but I don't understand how and why.

Will appreciate any help, thanks!
 
Physics news on Phys.org


You should know that a complex number can be represented as a vector in the gauss plane, where te x-axis is the Real axes and the y-axis is the Imaginary axis. Take for example the complex number z=a+ib (where a and b are real), it can be represented by the vector (a,b). You can use also a polar coordinate system where r is the modulus of the vector (a,b) and \phi is the so-called anomaly, meaning the angle between the vector and the positive side of the x-axis. So you have:

z=r\left(\cos(\phi)+i\sin(\phi)\right)

If you consider the taylor expansion of the function sine, cosine and , e^{x}you can observe that:

\cos(x)=\frac{e^{ix}+e^{-ix}}{2}

\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}

and so:

e^{ix}=\cos(x)+i\sin(x)

So you can write a complex number with Euler's notation:

z=re^{i\phi}

Since in your equation z=1, i.e. the vector (1,0), in Euler's notation you have r=1 and \phi can be 0, 2\pi i, 4\pi i and so on, since they represent the same angle (i.e sine and cosine are periodic and the period is 2\pi and so also the exponential is periodic, and the period is 2\pi i
 
Last edited by a moderator:


Thank you for the detailed explanation!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top