# why does i exp(it)=1

• Sep 8th 2011, 06:21 AM
blueyellow
why does i exp(it)=1
why does i exp(it)=1?
• Sep 8th 2011, 06:22 AM
alexmahone
Re: why does i exp(it)=1
Quote:

Originally Posted by blueyellow
why does i exp(it)=1?

It doesn't.

Do you mean when does i exp(it) = 1?
• Sep 8th 2011, 06:28 AM
blueyellow
Re: why does i exp(it)=1
no I'm just trying to make sense of something someone wrote:

= |∫(t = 0 to π) (iRe^(it) dt) / (1 + R^4 e^(4it))|
≤ ∫(t = 0 to π) R dt / |1 + R^4 e^(4it)|
• Sep 8th 2011, 06:31 AM
alexmahone
Re: why does i exp(it)=1
Quote:

Originally Posted by blueyellow
no I'm just trying to make sense of something someone wrote:

= |∫(t = 0 to π) (iRe^(it) dt) / (1 + R^4 e^(4it))|
≤ ∫(t = 0 to π) R dt / |1 + R^4 e^(4it)|

That's because |i exp(it)| = 1.
• Sep 8th 2011, 06:52 AM
blueyellow
Re: why does i exp(it)=1
why?
I tired to figure it out but I can't. I have an exam retake tomorrow and I really can't think why |i exp(it)|=1
so I'd appreciate it if you help
• Sep 8th 2011, 06:57 AM
alexmahone
Re: why does i exp(it)=1
Quote:

Originally Posted by blueyellow
why?
I tired to figure it out but I can't. I have an exam retake tomorrow and I really can't think why |i exp(it)|=1
so I'd appreciate it if you help

|i exp(it)| = |i||exp (it)| = 1*1 = 1
• Sep 8th 2011, 01:53 PM
mr fantastic
Re: why does i exp(it)=1
Quote:

Originally Posted by alexmahone
|i exp(it)| = |i||exp (it)| = 1*1 = 1

Just to elaborate: exp(it) = cos(t) + i sin(t) therefore |exp(it)| = cos^2(t) + sin^2(t) = 1. All of which should be known by the OP.