How Does Rotating a Complex Integral Prove an Inequality?

[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 \squareSo 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?

 

[Office_Shredder]

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|

 

[Zorba]

Excellent, thanks for the quick reply.

 

[Office_Shredder]

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