1. ## complex transformtion/mapping

can someone please solve the following problem;

Thanks

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

$w=z^2+z-3$

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

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

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

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

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

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

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

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

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

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

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

$w=z^2+z-3$

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

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

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

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

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

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

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

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

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

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

Thanks for the help