Complex number question

• Jul 30th 2009, 12:36 AM
Stroodle
Complex number question
If $\displaystyle z=x+yi$ find the values of $\displaystyle x$ and $\displaystyle y$ such that $\displaystyle \frac{z-1}{z+1}=z+2$

• Jul 30th 2009, 12:52 AM
red_dog
$\displaystyle \frac{x-1+yi}{x+1+yi}=x+2+yi\Rightarrow x-1+yi=x^2+3x-y^2+2+i(2xy+3y)\Rightarrow$

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

Now, you have to solve the system. Can you do that?
• Jul 30th 2009, 12:57 AM
Gamma
Quote:

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

$\displaystyle \frac{z-1}{z+1}=z+2\Rightarrow z-1=z^2+3z+2 \Rightarrow 0$$\displaystyle =z^2+2z+3 \Rightarrow z=\frac{-2\pm \sqrt {4-12}}{2}$$\displaystyle =\frac{-2\pm \sqrt {-4\cdot 2}}{2}=-1 \pm i\sqrt{2}$
• Jul 30th 2009, 12:59 AM
Stroodle
Awesome. THanks for your help guys
• Sep 5th 2009, 10:50 PM
Mr Smith
Quote:

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

How did you get from there-to-there, to-there?
• Sep 5th 2009, 10:57 PM
Prove It
Quote:

Originally Posted by Mr Smith
How did you get from there-to-there, to-there?

Multiplying both sides of the equation by $\displaystyle z + 1$.