1. ## complex transformtion/mapping

can someone please solve the following problem;

Thanks

2. $\displaystyle z=x+iy$
$\displaystyle w=u+iv$

$\displaystyle w=z^2+z-3$

$\displaystyle =(x+iy)^2+(x+iy)-3$

$\displaystyle =x^2+2ixy-y^2+x+iy-3$

$\displaystyle =(x^2-y^2+x-3)+i(2xy+y)$

So:
$\displaystyle u=x^2-y^2+x-3$ and $\displaystyle v=2xy+y$

(a) If $\displaystyle v=c$ then $\displaystyle 2xy+y=c$ or

$\displaystyle y=\frac{c}{2x+1}$

(b) $\displaystyle Re(z)=k$ so $\displaystyle Re(x+iy)=x=k$

So: $\displaystyle u=k^2-y^2+k-3$ and $\displaystyle v=y(2k+1)$

(c) For c=1: $\displaystyle y=\frac{1}{2x+1}$

For k=0: $\displaystyle u=-y^2-3$ and $\displaystyle v=y$ so $\displaystyle u=-v^2-3$

$\displaystyle z=x+iy$
$\displaystyle w=u+iv$

$\displaystyle w=z^2+z-3$

$\displaystyle =(x+iy)^2+(x+iy)-3$

$\displaystyle =x^2+2ixy-y^2+x+iy-3$

$\displaystyle =(x^2-y^2+x-3)+i(2xy+y)$

So:
$\displaystyle u=x^2-y^2+x-3$ and $\displaystyle v=2xy+y$

(a) If $\displaystyle v=c$ then $\displaystyle 2xy+y=c$ or

$\displaystyle y=\frac{c}{2x+1}$

(b) $\displaystyle Re(z)=k$ so $\displaystyle Re(x+iy)=x=k$

So: $\displaystyle u=k^2-y^2+k-3$ and $\displaystyle v=y(2k+1)$

(c) For c=1: $\displaystyle y=\frac{1}{2x+1}$

For k=0: $\displaystyle u=-y^2-3$ and $\displaystyle v=y$ so $\displaystyle u=-v^2-3$

Thanks for the help