Understanding the Euler Formula and its Application in Complex Analysis

[EternusVia]

Rudin makes the following statement in his Real and Complex Analysis, page 3:

If z = x + iy, x and y are real, then e^z = e^xe^iy. Hence |e^z| = e^x.

I don't understand what's happening here. If I draw e^z on coordinate axes with imaginary numbers on the y-axis and real numbers on the x-axis, I get |e^z| = √((e^x)² + (e^iy)². Does this equal e^x somehow?

 

[Ray Vickson]

Rudin makes the following statement in his Real and Complex Analysis, page 3:

If z = x + iy, x and y are real, then e^z = e^xe^iy. Hence |e^z| = e^x.

I don't understand what's happening here. If I draw e^z on coordinate axes with imaginary numbers on the y-axis and real numbers on the x-axis, I get |e^z| = √((e^x)² + (e^iy)². Does this equal e^x somehow?

No, you do NOT get what you claim when you plot the points correctly in the $(x,y)-$plane. In fact, $e^{x+iy} = e^x e^{iy} \neq e^x + i|e^{iy}|,$ so $|e^{x+iy}| \neq \sqrt{(e^x)^2 + (e^{iy})^2}.$ Anyway, $(e^{iy})^2$ need not be a positive real number---it could be negative or imaginary.

 

[EternusVia]

No, you do NOT get what you claim when you plot the points correctly in the $(x,y)-$plane. In fact, $e^{x+iy} = e^x e^{iy} \neq e^x + i|e^{iy}|,$ so $|e^{x+iy}| \neq \sqrt{(e^x)^2 + (e^{iy})^2}.$ Anyway, $(e^{iy})^2$ need not be a positive real number---it could be negative or imaginary.

You're right. I definitely made a mistake there. So how do we deduce $| e^z | = e^x$?

 

[Ray Vickson]

You're right. I definitely made a mistake there. So how do we deduce $| e^z | = e^x$?

PF rules do not allow me to tell you that. You must do it on your own.

What does your textbook say? Is there any relevant material in your course notes? Have you looked on-line?

 

[Mark44]

Moved thread, since this seems to be more of a conceptual question than a homework question.

 

[EternusVia]

PF rules do not allow me to tell you that. You must do it on your own.

What does your textbook say? Is there any relevant material in your course notes? Have you looked on-line?

I'm trying to work through some of Rudin's book on my own, so no course notes. But I think I might have something.

Rudin says that $|e^{it}|^2 = e^{it} * e^{-it} = e^0 = 1$ and we have $|e^{it}| = 1$. Applying this procedure to $|e^z|$ we have

$|e^z|^2 = e^z * \bar{ e^z } = e^{x + iy} * e^{x - iy} = e^{2x}$ and thus $|e^z| = e^x$.

Does that work?

 

[Mark44]

Looks OK to me, but I think you can do it with less work. e^it represents a point on the unit circle in the complex plane, so |e^it| = 1.

$|e^z| = |e^{x + iy}| = |e^x||e^{iy}| = |e^x| = e^x$

 

[EternusVia]

Looks OK to me, but I think you can do it with less work. e^it represents a point on the unit circle in the complex plane, so |e^it| = 1.

$|e^z| = |e^{x + iy}| = |e^x||e^{iy}| = |e^x| = e^x$

Ah! Of course. Thank you.

 

[Ray Vickson]

Ah! Of course. Thank you.

Also: look at the famous Euler formula: $e^{it} = \cos(t) + i \, \sin(t)$ for any real $t$. One easy way to see this is to look at the series expansion. By definition, for any quantity $w$ we have
$$e^w = 1 + w + \frac{1}{2!} w^2 + \frac{1}{3!} w^3 + \cdots + \frac{1}{n!} w^n + \cdots.$$
Apply this to $w = i t$ (where $t$ is real), and separate out the real and imaginary parts, to get
$$e^{it} = 1 + it + \frac{1}{2!} (it)^2 + \frac{1}{3!} (it)^3 + \frac{1}{4!} (it)^4 + \frac{1}{5!} (it)^5 + \cdots \\ = \left( 1 - \frac{1}{2!} t^2 + \frac{1}{4!} t^4 - \cdots \right) + i \left( t - \frac{1}{3!} t^3 + \frac{1}{5!} t^5 - \cdots \right) \\ = \cos(t) + i \, \sin(t)$$

Note that $|e^{it}|^2 = \cos^2(t) + \sin^2(t) = 1$ for any real $t$.

See, eg., http://mathworld.wolfram.com/EulerFormula.html