Find the image of the curve x^2-y^2=-2 under the mapping w=z^2
Find the image of the curve $x^2-y^2=-2$ under the mapping $w=z^2$.
Explanation of Method
Split $w=z^2$ into real and imaginary parts $u=x^2-y^2$, $v=2xy$. The curve's equation directly fixes $u$; showing $v=2xy$ takes every real value as the curve is traced identifies the image as a full vertical line.
Step 1 · Split $w=z^2$ into components
$$ w=(x+iy)^2=(x^2-y^2)+i(2xy) \quad\Longrightarrow\quad u=x^2-y^2,\ \ v=2xy $$
Step 2 · Apply the curve's equation
The curve is exactly $u=x^2-y^2=-2$, so:
$$ \boxed{u=-2} $$
Step 3 · Parametrize the hyperbola
$x^2-y^2=-2$ is a hyperbola opening along the $y$-axis (vertices at $(0,\pm\sqrt2)$). Parametrize with $x=\sqrt2\sinh t$, $y=\pm\sqrt2\cosh t$ for $t\in\mathbb R$ (the two signs trace the upper and lower branches; note $y^2-x^2=2\cosh^2t-2\sinh^2t=2$ ✓).
Step 4 · Compute $v$ along the parametrization
$$ v=2xy=2(\sqrt2\sinh t)(\pm\sqrt2\cosh t)=\pm4\sinh t\cosh t=\pm2\sinh(2t) $$
As $t$ ranges over all of $\mathbb R$, $\sinh(2t)$ ranges over all of $\mathbb R$ (it is a continuous, strictly increasing bijection $\mathbb R\to\mathbb R$), so $v$ takes every real value — on the upper branch alone (fixing the $+$ sign), and again on the lower branch.
Step 5 · Conclusion
$$ \boxed{w=z^2 \ \text{maps} \ \{x^2-y^2=-2\} \ \text{onto the full line} \ u=-2,\ v\in\mathbb R} $$
Verification
| $(x,y)$ on curve | $w=z^2$ |
|---|---|
| $(0,\sqrt2)$ (vertex) | $-2+0i$ |
| $(1,\sqrt3)$ | $(1-3)+i(2\sqrt3)=-2+2\sqrt3\,i$ |
| $(-1,\sqrt3)$ | $(1-3)+i(-2\sqrt3)=-2-2\sqrt3\,i$ |
Every sampled point has real part exactly $-2$, with the imaginary part varying continuously and unboundedly as the curve point moves — consistent with the image being the entire vertical line.
Result
Each branch of the "level-$(-2)$" hyperbola $x^2-y^2=-2$ maps onto the entire vertical line $u=-2$ — a hallmark of $w=z^2$ sending the two families of hyperbolas $x^2-y^2=\text{const}$ and $xy=\text{const}$ to the coordinate grid lines of the $w$-plane.