# Fundamental complex inequality

1. Jul 11, 2010

### Zorba

To proove the inequality:

$$\left | \int_a^b f(t) dt \right | \le \int_a^b | f(t) | dt$$

for complex valued $$f$$, use the following:

$$\textrm{Re}\left [ e^{-\iota\theta}\int_a^b f(t) dt\right ] = \int_a^b \textrm{Re}[e^{-\iota\theta}f(t)] dt \le \int_a^b | f(t) | dt$$

and then if we set:

$$\theta = \textrm{arg}\int_a^b f(t) dt$$

the expression on the left reduces to the absolute value $$\square$$

So the last part, where he says that it reduces to the modulus by using that substitution of theta, I don't get it... how does that reduce to the modulus?

2. Jul 11, 2010

### Office_Shredder

Staff Emeritus
Let $$z=\int_a^b f(t)dt$$

Let $$\theta=arg(z)$$

Then we can write $$z=Re^{i \theta}$$ where $$R=|z|$$

Now look at $$e^{-i \theta} z = e^{-i \theta} R e^{i \theta} = R = |z|$$

3. Jul 11, 2010

### Zorba

Excellent, thanks for the quick reply.

4. Jul 11, 2010

### Office_Shredder

Staff Emeritus
To give a bit more of an exposition, you can think of multiplying a complex number by $$e^{i \theta}$$ as being the same thing as rotating it in the complex plane by an angle of $$\theta$$. In this case what they did was rotate the number so that it lined up with the real axis

It's pretty clever; I've never seen this proof done like this before