2
$\begingroup$

Describe mathematically and draw what happens to the half-strip $R = \{z=x+iy: 0 \leq x \leq 1, y \geq 0 \}$ under the mapping $f(z) = z^2$

I need help with describing and drawing the mapping.

solution:

For, $z = x+iy$ and $w=f(z)=z^2$

$\Rightarrow w=(x+iy)^2 = (x^2-y^2) + 2xyi$.

So $w=u+iv \Rightarrow u=x^2-y^2$ and $v = 2xy$

case i: $u=x^2-y^2 = c_1, c_1 >0$

The graph in the $xy$-plane is the hyperbola cutting the $x$-axis. The $uv$-plane $u=c_1$ represents a vertical line. That is, hyperbolas in the $xy$-plane are mapped to vertical lines in the $uv$-plane.

caseii: $v = 2xy = c_2, c_2>0$

$v=c_2$ represents a horizontal line. The hyperbolas in the $xy$-plane are mapped to horizontal lines in the $uv$-plane.

drawing:

enter image description here

$\endgroup$
1
$\begingroup$

The strip is the thing that should be drawn on the $x-y$ plane and what it goes to on the $u-v$ plane. It's probably best to consider the boundary first. The first part is the positive $y$ axis with $x=0.$ This goes $iy\to (iy)^2 = -y^2$ so its image is the negative real axis. Next do the segment on the $x$ axis between $0$ and $1.$ This goes from $x\to x^2$ so its image is the same segment.

Last, the segment from $1$ to $1+i\infty.$ This goes $(1+iy)\to (1+iy)^2 = 1-y^2+2iy.$ This is a curve and we can let $t=2y$ so $t\in(0,\infty)$ and we have $v=t$ and $u=1-(t/2)^2.$ This is just the graph $u = 1-(v/2)^2$ for $v>0,$ so looks like a sideways half parabola, with vertex at $(1,0)$ and opening toward the negative real axis.

So the map takes the boundary of the strip to the this wedge shape with a striaght line bottom and parabolic top. I'll leave you to decide where the interior goes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.