1. ## Complex number question

If $z=x+yi$ find the values of $x$ and $y$ such that $\frac{z-1}{z+1}=z+2$

2. $\frac{x-1+yi}{x+1+yi}=x+2+yi\Rightarrow x-1+yi=x^2+3x-y^2+2+i(2xy+3y)\Rightarrow$

$\Rightarrow\left\{\begin{array}{ll}x^2+3x-y^2+2=x-1\\2xy+3y=y\end{array}\right.\Rightarrow\left\{\be gin{array}{ll}x^2-y^2+2x=-3\\2y(x+1)=0\end{array}\right.$

Now, you have to solve the system. Can you do that?

3. Originally Posted by Stroodle
If $z=x+yi$ find the values of $x$ and $y$ such that $\frac{z-1}{z+1}=z+2$

$\frac{z-1}{z+1}=z+2\Rightarrow z-1=z^2+3z+2 \Rightarrow 0$ $=z^2+2z+3 \Rightarrow z=\frac{-2\pm \sqrt {4-12}}{2}$ $=\frac{-2\pm \sqrt {-4\cdot 2}}{2}=-1 \pm i\sqrt{2}$

4. Awesome. THanks for your help guys

5. Originally Posted by Gamma
$\frac{z-1}{z+1}=z+2\Rightarrow z-1=z^2+3z+2$

How did you get from there-to-there, to-there?

6. Originally Posted by Mr Smith
How did you get from there-to-there, to-there?
Multiplying both sides of the equation by $z + 1$.