# Doubt on e^jwt

1. May 1, 2010

### palgun kumar

it is given that e^jwt=cos(wt)+jsin(wt)...
but j=1 with angle 90 degrees..now by multiplying any function by j means multiplying the magnitude by 1 and phase change of 90 degree...
so,
jsin(wt)=sin(wt)*1 with angle 90 degrees
= sin(wt+90)
=cos(wt)
so can we write e^jwt=cos(wt)+cos(wt)
=2cos(wt)

2. May 1, 2010

### D H

Staff Emeritus
j is not 1. It is the square root of -1.

3. May 1, 2010

### palgun kumar

square root of -1 is 1 with angle of 90 degrees or 1 with angle of -90 degrees

4. May 1, 2010

### Gerenuk

You could see it this way, if you like, but then you should be aware that sin(x) is composed of two parts with different phases! In particular

$$\sin(x)=\frac12\left(e^{i(x-\frac{\pi}{2})}+e^{i(-x+\frac{\pi}{2})}\right)$$

Now if you advances both phases you get
$$i\sin(x)=\frac12\left(e^{ix}+e^{i(-x+\pi)}\right)$$

Which is not
$$\sin(x+\frac{\pi}{2})=\frac12\left(e^{ix}+e^{-ix}\right)$$

So this rule of thumb rule you mention isn't really working.

The explanation is: Advancing phases in the sense you have learned is not the same as multiplying with complex numbers! That only works for pure exponentials.

5. May 2, 2010

### uart

No it's the square root of (-1).

Usually it's given the symbol "i" but in Engineering it's quite common to use the symbol "j" instead. The reason is because in Electrical Engineering it's very often used in AC circuit analysis where the symbol "i" is already used to denote electrical current.