why i=e^î*(pi/2) ?

**problady** · March 26th 2011, 09:00 AM

why i=e^î*(pi/2)

Guys please let us know that I am not so good in maths... and I think the question that I am asking is not so easy ... however I strongly believe that in this forum we have people who can make me understood this (although I am not so good in maths) and have abilities to understand this kind of questions! ... THANKS

**Phugoid** · March 26th 2011, 09:08 AM

Originally Posted by problady

Guys please let us know that I am not so good in maths... and I think the question that I am asking is not so easy ... however I strongly believe that in this forum we have people who can make me understood this (although I am not so good in maths) and have abilities to understand this kind of questions! ... THANKS

$e^{ix}$ is defined as $e^{ix} = \cos(x) + i \sin(x)$

Hence, for $x= \frac{\pi}{2}$ , we have...

$e^{i \frac{\pi}{2}} = \cos(\frac{\pi}{2}) = i \sin(\frac{\pi}{2})$

If you know your trigonometry, then you'll know that...

$\cos(\frac{\pi}{2}) = 0$

... and...

$\sin(\frac{\pi}{2}) = 1$

Hence, the equation becomes...

$e^{i\frac{\pi}{2}} = 0 + i (1) = i$

**TheChaz** · March 26th 2011, 09:12 AM

Euler's formula states that
$e^{i \theta} = cos(\theta) + i*sin(\theta)$

When theta is pi/2, the cosine term vanishes and the sine term is 1.

**Phugoid** · March 26th 2011, 09:12 AM

Originally Posted by problady

Guys please let us know that I am not so good in maths... and I think the question that I am asking is not so easy ... however I strongly believe that in this forum we have people who can make me understood this (although I am not so good in maths) and have abilities to understand this kind of questions! ... THANKS

$e^{ix}$ is defined as $e^{ix} = \cos(x) + i \sin(x)$

Hence, for $x= \frac{\pi}{2}$ , we have...

$e^{i \frac{\pi}{2}} = \cos(\frac{\pi}{2}) = i \sin(\frac{\pi}{2})$

If you know your trigonometry, then you'll know that...

$\cos(\frac{\pi}{2}) = 0$

... and...

$\sin(\frac{\pi}{2}) = 1$

Hence, the equation becomes...

$e^{i\frac{\pi}{2}} = 0 + i (1) = i$

**TheChaz** · March 26th 2011, 09:23 AM

You've done this a few times on threads to which I have contributed. (sandwiched my post with duplicates of yours)
I don't mean to be redundant- it's kinda slow typing laTex on an iPhone.

Anyhow, I can't think of a legitimate reason why you would create a^n exact duplicate post, then edit (almost delete) your previous post to ".".

It's just obnoxious in my book
Regardless, the question is answered.

**Phugoid** · March 26th 2011, 09:57 AM

Originally Posted by TheChaz

You've done this a few times on threads to which I have contributed. (sandwiched my post with duplicates of yours)
I don't mean to be redundant- it's kinda slow typing laTex on an iPhone.

Anyhow, I can't think of a legitimate reason why you would create a^n exact duplicate post, then edit (almost delete) your previous post to ".".

It's just obnoxious in my book
Regardless, the question is answered.

Please, don't jump to assumptions about my intentions.

There seems to be a problem with this forum/my computer, whereby once I hit the post reply button, the post is submitted, but the screen returns to the 'Go Advanced' page, leading me to believe that the post hasn't been submitted, thus leading to me posting it again.

So, rather than being obnoxious, I am in fact the victim of some sort of error.

And not sure what you mean about the "."... the first I knew that I had made a duplicate post was when I returned after being e-mailed about your reply, and I haven't edited it or touched it since.

**HallsofIvy** · March 27th 2011, 05:58 AM

A little more detail.

Since we know how to add and multiply complex numbers, a good way to extend other functions from real numbers to complex numbers is to use their Taylor series.

The Taylor series for $e^x$ is
$\sum_{n=0}^\infty \frac{x^n}{n!}= 1+ x+ (1/2)x^2+ (1/3!)x^3+ \cdot\cdot\cdot$

The Taylor series for $cos(x)$ is
$\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$

and the Taylor series fror $sin(x)$ is
$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$ .

If you replace x with ix in the Taylor series for $e^x$ , you get
$1+ ix+ \frac{-1x^2}{2!}+ \frac{-ix^2}{3!}+ \cdot\cdot\cdot$

Because $i^0= 1$ , $i^1= i$ , $i^2= -1$ , $i^3= -i$ , and $i^4= 1$ , every odd power involves i while every even power does not. Separating real and imaginary parts,
$e^{ix}= (1- \frac{1}{2}x^2+ \frac{1}{4!}x^4- \cdot\cdot\cdot)+ i(x- \frac{1}{3!}x^3+ \frac{1}{5!}x^5-\cdot\cdot\cdot)$ $= cos(x)+ i sin(x)$

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