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.


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.


enter image description here


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.


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