How to solve a complex equation using quadratic formula

AI Thread Summary
To solve the complex equation 3(x^2 + y^2) + (x - iy)^2 + 2(x + iy) = 0, the discussion emphasizes converting the equation into the form f(x, y) + i g(x, y) = 0. Participants suggest using the quadratic formula after simplifying the equation by expanding and grouping real and imaginary parts. It is recommended to define z = x + iy and its conjugate to create two equations that can be solved simultaneously. The conversation highlights the importance of understanding complex conjugates and their role in finding solutions. Ultimately, the goal is to derive the correct values for x and y that satisfy the original equation.
username12345
Messages
47
Reaction score
0

Homework Statement



Find all complex solutions to the following equation:

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

Homework Equations



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

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.
 
Physics news on Phys.org
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??
 
praharmitra said:
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...

I don't understand this.

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

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?
 
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
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top