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The question is:
Consider the transformation
$$w=z^2+3z$$ from the $z$-plane where $z=x+iy$ and $w=u+iv$.

Determine the image of the line $y=1$ and $x=3$ in the $w$-plane

My attempt at a solution: $$\begin{align} u+iv=w&=z^2+3z\\ &=(x+iy)^2+3(x+iy)\\ &=x^2+ixy+ixy+(iy)^2+3x+3iy\\ &=x^2+2ixy-y+3x+3iy\\ &=x^2+3x-y+i(2xy+3y) \end{align}$$ So $$\begin{align} &u=x^2+3x-y\\ &v=2xy+3y \end{align}$$

$$\begin{align} w&=z^2+3z\\ z^2+3z-w&=0\\\text(Which is a Quadratic euation)\\\\ x+iy=z&=\frac{-3\pm\sqrt{9-4w}}{2}\\ &=\frac{-3\pm\sqrt{9-4(u+iv)}}{2} \end{align}$$ So how do I now break $u$ and $iv$ appart?

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up vote 1 down vote accepted

Equation $$ w=z^2+3z $$ is a quadratic equation w.r.t. $z:$ $$z^2+3z-w=0 \\ z_{1,2}=\frac{-3\pm\sqrt{9-4w}}{2} $$ Image of the line $\gamma_1=\{z=x+iy:\quad y=1\}$ can be obtained by substitution $y=1$ in $$\begin{align} &u=x^2+3x-y\vert_{y=1}=x^2+3x-1\\ &v=2xy+3y\vert_{y=1}=2x+3 \; \Rightarrow \;x=\frac{v-3}{2} \end{align}$$ hence $$u=\left(\frac{v-3}{2}\right)^2+3\cdot\frac{v-3}{2}-1 \, .$$ Analogously, image of another line $\gamma_2=\{z=x+iy:\quad x=3\}$ \begin{align} &u=x^2+3x-y\vert_{x=3}=18-y \; \Rightarrow \;y=18-u \\ &v=2xy+3y\vert_{x=3}=6y+3y=9y \; \Rightarrow \;v=9(18-u)=-9u+162 \end{align} hence $$v=-9u+162$$

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Why substitute $y=1$ and not $x=3$? – Gineer Nov 9 '12 at 6:56

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