# Complex number exponential subtraction

1. Feb 1, 2005

### Exulus

Hi all,

Im having a bit of trouble with a question. I have to convert:

$$Ke^{j\delta} - Ke^{j\psi}$$

Into the form

$$re^{j\theta}$$

This is the second part of the question, the first part was an addition instead of subtraction which i managed by using this formula:

$$z_1 + z_2 = K(e^{j\delta} + e^{j\psi}) = Ke^{j(\delta + \psi)/2}(e^{j(\delta - \psi)/2} + e^{-j(\delta - \psi)/2}) = 2K\cos((\delta - \psi)/2).e^{j(\delta + \psi)/2}$$

I cant really see where to go with the subtraction though...is it maybe to do with a sin rule? To be honest i dont fully understand the formula above but it was given to us...Ive fiddled around with the maths for a while but its totally headbanging :( Hoping someone can help!

2. Feb 1, 2005

### HallsofIvy

Since you already have a formula for sums, how about using the fact that $e^{i(\psi+ \pi)}= e^{i\psi}e^{i\pi}= -e^{i\psi}$? In other words, replace $\psi$ with $\psi+ \pi$ in your formula.

3. Feb 1, 2005

### dextercioby

Okay.It's not difficult.Jst follow the same pattern using the half angle

$$D=z_{1}-z_{2}=Ke^{j(\frac{\delta+\psi}{2})}[e^{j(\frac{\delta-\psi}{2})}-e^{j(\frac{\psi-\delta}{2})}]=Ke^{j(\frac{\delta+\psi}{2})}[e^{j(\frac{\delta-\psi}{2})}-e^{-j(\frac{\delta-\psi}{2})}]=[2jK\sin(\frac{\delta-\psi}{2})]e^{j(\frac{\delta+\psi}{2})}$$

Okay?

Daniel.

4. Feb 1, 2005

### Exulus

brilliant thanks for your help! :)

5. Apr 25, 2006

### ph23ms

URGENT: Can anyone help me with this?

Can anyone tell me what difference it makes to the subtraction/addition formulae when the base factors are not the same? i.e

[TEX]D=z_{1}-z_{2}=Ke^{j(\delta)}-Le^{j(\psi)}[/TEX]

I have fiddled with the above formulae, but cant seem to get anything sensible looking from it... And I need to solve this quite urgently!!!