# COMPLEX SOLUTIONS help!

1. Apr 27, 2008

### tizzful

First problem
z^2+2z+1=0
where the z in 2z is the conjugate (has a little line ontop)
I just ignored the conjugate because I wasn't sure how to solve it, and I got -1 which is one of the solutions but there's also 1+2i and 1-2i which I understand because they're both conjugate of each other but I don't understand how they got it.
Second problem
z^3-3z^2+4z-12=0 given 2i is a solution... I don't even understand what they mean.

2. Apr 27, 2008

### Mr.Slava

As I see your problem is
$$z^{2} + 2 z^{*} + 1 = 0$$

if you will search solution in following form $$z = a + i b$$, $$a, b$$ are both real numbers and insert it to your main equation then you will have system of two simple algebraic equations under $$a, b$$ and you'll find $$a = 1, b = \pm 2$$. Solve your second problem in the same way and get the answer.

3. Apr 27, 2008

### tizzful

I get that z=a+ib but how did you get values for a and b? I feel really stupid asking this but I dont see it. I tried solving it and then making b=0 and a=0 and I'm not getting 1 and 2 as values...

4. Apr 27, 2008

### Mr.Slava

What are your equations for a and b?

5. Apr 27, 2008

### HallsofIvy

Staff Emeritus
Do the algebra. If z= a+ ib, then $\overline{z}$= a- ib so $z^2+ 2\overline{z}+ 1= (a+ ib)^2+ 2(a- ib)+ 1= 0$. Separate the real and imaginary parts and you have two equations for a and b.

6. Apr 27, 2008

### tizzful

Yeah I got that far but i'm not sure how to seperate real and imaginary parts. Is it literally just placing all the real parts and making them equal to 0 and all the imaginary parts and make them equal to 0? Sorry our lecturer didn't go through this and so I'm just left lost. :shy:
Thank you

7. Apr 27, 2008

### HallsofIvy

Staff Emeritus
Yes, it literally is! If a+ bi= c+ di, then a= c and b= d. That's part of the definition of "complex number".