complex numberscomplex-analysisfunctions
I have a difficulty understanding the basics of complex functions. My exercise looks like this:
"The $z$-plane region $D$ consists of the complex numbers $z = x + yi$ that satisfy the given conditions:
$$x + y = 1, w = \bar{z}$$
Describe the image $R$ of $D$ in the $w$-plane under the given function $w = f(z)$."
I just really don't know how to tackle this exercise, I know it's basic but any suggestions on how to go at it would be appreciated.
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 $x=4-(y/4)^2$.
HINT
Let's tackle the first one. A similar approach is required for the other two.
For $z=x+iy$, $x,y\in \mathbb{R}$ we get $$|z-1+i|=1\Rightarrow\\|x-1+i(y+1))|=1\Rightarrow\\\sqrt{(x-1)^2+(y+1)^2}=1\Rightarrow\\(x-1)^2+(y+1)^2=1$$
Does this-hopefully it does-remind you a more general equation of a circle?
Let's also look at the third one as well.
We have $$\operatorname{Re}\Big(\frac{z+1}{z-1}\Big)=\Re\Big(\frac{z+1}{z-1}\Big)=\Re\Big[\frac{(z+1)(\bar{z}-1)}{(z-1)(\bar{z}-1)}\Big]=\Re\Big[\frac{(z+1)(\bar{z}-1)}{|z-1|^2}\Big]\gt1$$ Now for $z=x+iy$, $x,y\in \mathbb{R}$ you can substitute on the last relationship and obtain an equation for $x,y$-an inequality to be precise.
Best Answer
$$x+y=1\iff y=1-x$$
and you have the straight line $\;y=1-x\;$ in the complex plane, which you can also express as the set
$$\{z\in\Bbb C\;:\;\;z=x+(1-x)i\;,\;\;x\in\Bbb R\}$$
If you take a general element of this set and apply on it the transformation $\;w\;$ ,we get
$$w(x+(1-x)i):=x-(1-x)i$$
and the image is the straight line $\;y=-(1-x)=x-1\;$.