# Absolute Value of e(i)

1. Aug 27, 2010

### Vectorspace

Why does $$|e^i| = 1$$ ?

2. Aug 27, 2010

### CompuChip

For any complex number z, we define |z| as z z*, where z* is the complex conjugate of z, right?

Then seeing that
$$|e^{\mathrm{i} x}| = 1$$
for any real number x is a matter of applying the definition and simple algebraic rules.

3. Aug 27, 2010

### Vectorspace

Thank you very much buddy.
:-)

4. Aug 27, 2010

### nonequilibrium

How is it obvious that $$\left( e^{ix} \right) ^{*} = \left( e^{-ix} \right)$$? I'd use the sin/cos represenatation, but the way you suggested it implies there's a more direct way?

5. Aug 27, 2010

### CompuChip

Vectorspace, you're welcome.

Using the sin/cos representation makes it explicit, because then you split it into its real and imaginary part.
For myself I always use the "shortcut": the complex conjugate is obtained by replacing every i in the expression by -i.

6. Aug 28, 2010

### nonequilibrium

Oh I see. Is that shortcut something that needs proving, or is it something evident?

7. Aug 28, 2010

### Dickfore

The analytic continuation of the exponential function which preserves the fundamental property of the exponentials:

$$\exp{(z_{1} + z_{2})} = \exp{(z_{1})} \cdot \exp{(z_{2})}$$

and is equal to the natural exponential function on the real line, i.e.:

$$\exp{(x)} \equiv e^{x}, x \in \mathbb{R}$$

is given by:

$$\exp{(z)} \equiv \exp{(x + i y)} = e^{x} \, \left(\cos{(y)} + i \, \sin{(y)}\right)$$

You can show explicitly that this function:
1. It satisfies the above functional equation;

2. It is analytic everywhere on the (finite) complex plane by seeing if the Cauchy Riemann conditions are satisfied and that the partial derivatives are continuous;

3. It reduces to $e^{x}$ when $y = 0$ which is trivial.

Then, you simply use the definition of absolute value to show that:

$$|\exp{(z)}| = \sqrt{u^{2}(x, y) + v^{2}(x, y)} = \sqrt{e^{2 x} \, \cos^{2}{(y)} + e^{2 x} \, \sin^{2}{(y)}} = e^{x} = e^{\Re{z}}$$

8. Aug 28, 2010

### Dickfore

BTW, what 'mr. vodka' was asking. It is not necessarily true that:

$$f(\bar{z}) = \overline{f(z)}$$

See Schwartz reflection principle for further discussion.

9. Aug 28, 2010

### nonequilibrium

Thank you Dickfore. Could it be true though that your invalid formula is true in the special case of f being real and analytical? Then you can write the infinite sum and it seems reasonable that the complex conjugate of an infinite sum is the infinite sum of the complex conjugates, although I don't actually know how to prove that if $$\sigma = \sum_i^\infty a_i z^i$$ with a_i real and z complex, that $$\rho_n = \sum_i^n a_i \overline{z^i} \to \overline{\sigma}$$.

10. Aug 28, 2010

### Dickfore

The only real and analytic function is a trivial constant.

Proof:
$$w = f(z) = u(x, y) + i \, v(x, y), \; z = x + i \, y$$

$$w \in \mathbb{R} \Rightarrow v \equiv 0$$

From the Cauchy Riemann conditions:

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = 0$$

$$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial y} = 0$$

Because both partial derivatives of u w.r.t. x and y are identically equal to zero in the region of analyticity, it means $u = \mathrm{const.} \in \mathbb{R}$. Q.E.D.

11. Aug 28, 2010

### nonequilibrium

I'm sorry, I'm not familiar with complex analysis. With "real analytic function" I meant a function whereof the expansion $$\sum a_i z^i$$ has only real a_i's (maybe a characterisation of such a function is an analytical function with f(R) a set of reals?)