# Finding Real and Imaginary

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• Aug 27th 2008, 07:10 AM
iphysics
Finding Real and Imaginary
$m*z^2 - D*z +k = 0
$

Need to find the values of z in terms of real and imaginary. m, D and k are all constants. Help would be much aprecciated!
• Aug 27th 2008, 07:16 AM
Moo
Hello,
Quote:

Originally Posted by iphysics
$m*z^2 - D*z +k = 0
$

Need to find the values of z in terms of real and imaginary. m, D and k are all constants. Help would be much aprecciated!

Just find the discriminant !

$\Delta=D^2-4km$

The solutions are thus $z=\frac{D \pm \sqrt{D^2-4km}}{2m}$

If $D^2-4km <0$, we can write $D^2-4km=(-1)*\underbrace{(4km-D^2)}_{>0}=i^2 \cdot (4km-D^2)$

$\implies z=\frac{D \pm i ~\sqrt{4km-D^2}}{2m}=\frac{D}{2m} \pm i ~\frac{\sqrt{4km-D^2}}{2m}$

What are the real and imaginary parts ? (Tongueout)
• Aug 27th 2008, 07:20 AM
iphysics
Thanks for that man. Im going to scan the document in another topic, because it is near impossible for me to do. Will upload ASAP

BTW, The real component is = to $D/2m$. Dont know how to explain the negative sign!?
• Aug 27th 2008, 08:46 AM
mr fantastic
Quote:

Originally Posted by Moo
Hello,

Just find the discriminant !

$\Delta=D^2-4km$

The solutions are thus $z=\frac{{\color{red}-(-D)} \pm \sqrt{D^2-4km}}{2m}$

If $D^2-4km <0$, we can write $D^2-4km=(-1)*\underbrace{(4km-D^2)}_{>0}=i^2 \cdot (4km-D^2)$

$\implies z=\frac{{\color{red}-(-D)} \pm i ~\sqrt{4km-D^2}}{2m}=\frac{{\color{red}D}}{2m} \pm i ~\frac{\sqrt{4km-D^2}}{2m}$

What are the real and imaginary parts ? (Tongueout)

Quote:

Originally Posted by iphysics
Thanks for that man. Im going to scan the document in another topic, because it is near impossible for me to do. Will upload ASAP

BTW, The real component is = to $D/2m$. Dont know how to explain the negative sign!?

The explanation (in red) is simple ....

By the way, I think you'll have better luck with your other post if you type it out ... People like myself rarely open unknown files, for obvious reasons.