Please teach me how to solve this question...

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.

I assume you are using the * to represent conjugate. You should know that \displaystyle \begin{align*} w\,\overline{w} = |w|^2 \end{align*}, so we have \displaystyle \begin{align*} w\,\overline{w} = z + 6 \end{align*} and \displaystyle \begin{align*} \overline{w} = z - 2i \end{align*}. Subtracting the second equation from the first gives

\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 \begin{align*} w = 2 + 2i \end{align*} or \displaystyle \begin{align*} w = -1 + 2i \end{align*}