Transform $\Re(z)=1 \space , \Re(z)=\Im(z) \space and \space \Re(z)=-\Im(z)$ using the mapping $w=iz^2$

Question

Can someone please verify whether i am doing this the right way -Thanks.

$$w=iz^2=i(x^2+2xiy-y^2)=-2xy+i(x^2-y^2)$$

$$\color{green}{u=-2xy \tag{1}}$$ and $$\color{green}{v=x^2-y^2 \tag{2}}$$

$\color{green}{1. \space Line \space \Re(z)=x=1}$

Substituting $x=1 \space in \space (1)$ we get: $u=-2y \tag{i}$

$\Re(z)$ is on $x$ axis and therefore $y=0$ $\implies u = 0$.

$$(i) \implies \color{green}{y=-\dfrac{u}{2}}$$

Substituting $x=1 \space in \space (2)$ we get: $v=1-y^2\tag{ii}$ using $(a)$ we get $v=1-\dfrac{u^2}{4}$

$$\color{green}{\therefore \Re(z)=1 \rightarrow u=0 \space and \space v=1-\dfrac{u^2}{4}}$$

$\color{green}{2. \space Line \space \Re(z)=\Im(z) \implies y=x}$

for $x=y$ we get the following from $(1)$: $$u=-2x^2 \space \space or \space \space u=-2y^2$$

$$\implies u \le 0$$

for $x=y$ we get the following from $(2)$: $$v=0 $$

$$\color{green}{\therefore \Re(z)=\Im(z) \rightarrow u \le 0 \space and \space v=0}$$

$\color{green}{3. \space Line \space \Re(z)=\Im(z) \implies y=x}$

$v=0$ and $u \ge 0$

$$\color{green}{\therefore \Re(z)=-\Im(z) \rightarrow u \ge 0 \space and \space v=0}$$

EDIT:

how does the transformed plot look like, is it just all the points $(u = -\infty ; v = 0) \space and \space (u = \infty ; v = 0) $ i.e. straight line on real axis and a parabola as in the picture below?

Answer 1

2. and 3. look fine, but 1. has some small (notation) error(s):

You write:

• "...$\operatorname{Re}(z)$ is on $x$-axis and therefore $y=0$ ..."

  What do you mean here?

• $u=0$ is not true. Since $y\in \mathbb{R}\Rightarrow u =-2y \in \mathbb{R}$.

  Neither is this consistent with the picture you drew. (in that curve $u$ clearly travels through $\mathbb{R}$)

Aside from the small notation errors I find it difficult to read, because you use all those references. I would write it as follows:

$$\operatorname{Re} z = 1 \leftrightarrow z(y) = 1+iy \qquad y\in \mathbb{R}$$

Then:

$$\begin{align} w(z) = iz^2 &= i(1+iy)^2\\ & = i(1+2iy-y^2)\\ &= -2y +i (1-y^2)\\ \end{align}$$

Meaning $$u= -2y\qquad v=1-y^2$$

It's easier to work with parameterizations. The same is valid for 2. and 3. As an example

$$\operatorname{Re}z = \operatorname{Im} z \leftrightarrow z(x) = x+ix \qquad x \in \mathbb{R}$$

The plots look good.