Let $\displaystyle f(z)=z^2$

Find the images of the line $\displaystyle Re(z)=0$ and $\displaystyle Im(z)=1$ under the above mapping, $\displaystyle w=f(z)$

Just to make sure I'm understanding this curve sketching

Represented parametrically, $\displaystyle Re(z)=0$ is $\displaystyle z_1(t)=it$ and $\displaystyle Im(z)=1$ is $\displaystyle z_2(t)=t+i, t\in{\mathbb{R}}$

So $\displaystyle f(z_1(t))=-t^2+0i$ and $\displaystyle f(z_2(t))=t^2-1 +2ti$

$\displaystyle x_1=-t^2, x_2=t^2-1$

$\displaystyle y_1=0, y_2=2t$

$\displaystyle x_2=\frac{y^2}{4}-1$

Thus the line $\displaystyle Re(z)=0$ is mapped onto the negative portion of the x-axis and the line $\displaystyle Im(z)=1$ is mapped onto this parabola:$\displaystyle x_2=\frac{y^2}{4}-1$

Any mistakes present?