Roots of a complex equation

1. Mar 20, 2009

1. The problem statement, all variables and given/known data

Find all complex solutions to the following equation:

$$3(x^2 + y^2) + (x - iy)^2 + 2(x + iy) = 0$$

2. Relevant equations

I want to use the quadratic formula, but not sure if it applies here.

3. The attempt at a solution

This is as far as I can get. What I would like is some idea as to what technique to solve this.

2. Mar 20, 2009

praharmitra

I could give you two ways of solving this problem. I ask u the following questions first..

1. What is the solution to x + iy = 0, or say x+1 + i(y-1) = 0??? If you have answered these correctly, then I guess you should go ahead and convert your problem to a

f(x, y) + i g(x,y) = 0 type, and solve it...

2. Write z = x + iy. Thats one equation. Conjugate it. Thats another equation. Can u solve them together now??

3. Mar 21, 2009

I dont understand this.

Are you suggesting that the complex conjugates are roots?

The equation I supplied was expanded where x + iy was originally z and x - iy was "zed bar".

Does anyone know a procedure to follow to solve this?

4. Mar 21, 2009

praharmitra

k, i'll simplify it slightly more for you...

For method 1.

I want you to use the formula (a+b)^2 and open up every bracket... then club together all real terms together, and all imaginary terms (with an i) together...

So, now do u get something like f(x,y) + i g(x,y) = 0 ??? Can you tell me how to solve such a problem??

If no, then I'll just give one hint.... (x+1) + i ( y-1) = 0 is of the above form with f(x,y) = x+1 and g(x,y) = y-1.

The only solution to the above is (-1,1). can u tell me why?

For method 2.

ok, this i will solve slightly so that i can explain clearly....

put x + iy = z. then x - iy = z*

The equation becomes....

3 z z* + (z*)^2 + 2 z = 0

Now conjugate the above equation. we get

3 z* z + z^2 + 2 z* = 0

You now have two equations with variables z and z*. Solve for them..

Hint - Subtract them