# Simple Analysis Question

1. Mar 21, 2015

### EternusVia

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

If z = x + iy, x and y are real, then ez = exeiy. Hence |ez| = ex.

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

2. Mar 21, 2015

### Ray Vickson

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.

3. Mar 21, 2015

### EternusVia

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

Last edited: Mar 21, 2015
4. Mar 21, 2015

### Ray Vickson

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?

5. Mar 21, 2015

### Staff: Mentor

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

6. Mar 21, 2015

### EternusVia

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?

7. Mar 21, 2015

### Staff: Mentor

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

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

8. Mar 21, 2015

### EternusVia

Ah! Of course. Thank you.

9. Mar 21, 2015

### Ray Vickson

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