
Please help me!! >.<
Please teach me how to solve this question...(Bow)
Two complex number w and z are such that w*= z 2i and w^2=z+6. By eliminating z, find w in the form a+ib, where a and b are real and positive.
please help me! I really have many problems with complex number!!(Crying)

Re: Please help me!! >.<
I assume you are using the * to represent conjugate. You should know that $\displaystyle \displaystyle \begin{align*} w\,\overline{w} = w^2 \end{align*}$, so we have $\displaystyle \displaystyle \begin{align*} w\,\overline{w} = z + 6 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \overline{w} = z  2i \end{align*}$. Subtracting the second equation from the first gives
$\displaystyle \displaystyle \begin{align*} w\,\overline{w}  \overline{w} &= 6 + 2i \\ \overline{w}\left( w  1 \right) &= 6 + 2i \\ \left( a  b\,i \right) \left( a  1 + b\,i \right) &= 6 + 2i \\ a(a  1) + ab\,i  (a  1)b\,i  b^2 i^2 &= 6 + 2i \\ a^2  a + b^2 + b\,i &= 6 + 2i \\ a^2  a + b^2 = 6 \textrm{ and } b &= 2 \\ a^2  a + 4 &= 6 \\ a^2  a  2 &= 0 \\ (a  2)(a + 1) &= 0 \\ a = 2 \textrm{ or } a &= 1 \end{align*}$
So the solution is $\displaystyle \displaystyle \begin{align*} w = 2 + 2i \end{align*}$ or $\displaystyle \displaystyle \begin{align*} w = 1 + 2i \end{align*}$

Re: Please help me!! >.<
Are you sure you copied the problem correctly?
Assuming you did:
1. What does w^2=z+6 say about the complex number z? (Hint: what kind of values can w take?)
2. Do you know "eliminate z" algebraically? (If not, then that kind of algebra is something you should heavily review  it's very important.)
3. Just so you can see how you're doing  it's eventually a quadratic you'll need to solve, and so you'll have two solutions for w. And they'll be clean, like (integer) + i(integer).