Finding roots of a complex equaitions

emailzul_at_starhub.net.sg
Date: 01/12/05 

Date: 12 Jan 2005 06:45:32 -0800

Prove that the usual formula solves the quadratic equation

az^2+bz+c=0

where the coefficients a,b,c are complex numbers.
How do I start the prove?

What I did is simply do the complete the square method - can that be
considered a proof?
I obtain the following:

z= (-b plusminus (b^2-4ac)^1/2 ) / 2a

However, I checked books by James Ward Brown on Complex Variables and
he omit the minus sign(-) after the -b so he has

z= (-b PLUS (b^2-4ac)^1/2 ) / 2a only

I am not sure whether it is a printing error or there is a reason
behind this.
Lastly the answers to roots were given as

(-1+1/root2 ) + i/root2 and (-1-1/root2) - i/root2

I managed to obtain
-1 + root i and -1 - root i

Although it can be further simplified i don't see a way i can get the
denominator of the i to be root 2.

Any help is appreciated.
regards

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