0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ b) $v^2=16 (4-u)$ $\endgroup$ – Lozenges Sep 25 '16 at 5:59
1
$\begingroup$

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 $x=4-(y/4)^2$.

enter image description here

$\endgroup$
3
  • $\begingroup$ @mint Thank you for improving the latex formulas ! $\endgroup$ – Jean Marie Sep 25 '16 at 17:36
  • $\begingroup$ 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/… $\endgroup$ – mint Sep 25 '16 at 17:48
  • $\begingroup$ I will have have a look. I have added a (geogebra) figure to my answer. $\endgroup$ – Jean Marie Sep 25 '16 at 17:59

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.