# Finding image under complex transformation

I am trying to find the image of the following under the complex mapping where $z=x+iy$ and $w=u+iv$

a) Find the image of the first quadrant of z plane under $w=z^2$.

b) Find the image of the region $D$ bounded by $x=2, y=0,x^2-y^2=1,x\ge 0,y\ge 0$ under the transformation $w=z^2$.

I did so far

a) The image will be the upper half plane (1st quadrant + 2nd quadrant)

b) For this I'm not so sure. I think the image will be the region bounded by $v=0, u=1, v=4\sqrt{3}$ and the parabola $v^2=16(u-4)$.

Any help would be greatly appreciated. Thanks in advance.

• b) $v^2=16 (4-u)$ – Lozenges Sep 25 '16 at 5:59

For a) you are right.

For b), this is a region included in the first quadrant bounded by two line segments and a parabolic arc obtained resp. in the following way:

• the image of the line segment from $(1,0)$ to $(2,0)$ is clearly the line segment from $(1,0)$ to $(4,0)$.

• the image of the hyperbola arc $x^2-y^2=1$ starting in $(1,0)$ and ending in $(2,\sqrt{3})$ will be obtained easily by parametrizing the hyperbola as the set of $z$ of the form $z=\cosh(t)+i \sinh(t)$. In this way $$z^2=(\cosh(t)^2-\sinh(t)^2)+2i \sinh(t)\cosh(t)=1+\sinh(2t)i.$$ It is thus the line segment starting in $(1,0)$ and ending in $(1,4\sqrt{3}).$

• the image of vertical segment from $(2,0)$ to $(2,\sqrt{3})$ is the image of points $z=2+it \ \ (0 \leq t \leq \sqrt{3})$, i.e., $z^2=(4-t^2)+(4t)i$, i.e., the arc of curve $y=4t, x=4-t^2$ which is a parabolic arc (with horizontal axis) of the parabola with equation $x=4-(y/4)^2$.

• @mint Thank you for improving the latex formulas ! – Jean Marie Sep 25 '16 at 17:36
• That was nothing. Thank you very much for helping me so much. Can you help me with my following question math.stackexchange.com/questions/1940464/… – mint Sep 25 '16 at 17:48
• I will have have a look. I have added a (geogebra) figure to my answer. – Jean Marie Sep 25 '16 at 17:59