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Suppose $u=\frac{z(1+i)-i}{z+1}$ as $z\in \mathbb{C} \setminus\{-1\}$

What is the set of points $M(z)$ for which $u$ is a real number?

What is the set of points $M(z)$ for which $u$ is pure imaginary number?

What is the set of points $M(z)$ for which $|u|=\sqrt{2}$?

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How far did you get? What approaches did you try? Where did you get stuck? Why does this problem interest you? – Gerry Myerson Feb 18 '13 at 12:17
I tried to find the value of z for which $z=\overline{z}$ – pourjour Feb 18 '13 at 12:19
in general I need an algorithm or an approache to solve such problems – pourjour Feb 18 '13 at 12:19
In general, the brute force way of solving problems like $z = \bar z$ and $\mathop{Im}(z) = i\mathop{Re}(z)$ (I had that one on an exam on my first calculus course) is to rewrite $z$ as $x + iy$, with $x$ and $y$ real numbers, and go from there. As long as the original equations aren't too dirty, the resulting ones might be quite solvable. – Arthur Feb 18 '13 at 12:42
up vote 1 down vote accepted

My first thought.

Simplify $u$ to $1+i-\frac{1+2i}{z+1}$ then put $z=a+bi$ in the $u$ and you will get expression (after simplifying) $w=f_1\left(a,b\right)+f_2\left(a,b\right)i$.

1) $f_2(a,b)=0$
2) $f_1(a,b)=0$
3) $f_1^2(a,b)+f_2^2(a,b)=2$

Solve each case.

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There's possibly some geometrical solution. – zaarcis Feb 18 '13 at 12:29
I think the set of points for the first two question is a circle – pourjour Feb 18 '13 at 12:52
Yes, they are. They can be described by equations like this $\vert z-z_0 \vert = r$ – zaarcis Feb 18 '13 at 16:36

For the first two, start by writing $z=x+iy$, rewrite $u$ as $v+iw$ for some real $v,w$, then figure out appropriate conditions on $x,y$. For the third, you set $v^2+w^2=2$, and then figure out appropriate $x,y$ conditions.

Now, the map $z\mapsto u(z)$ is what is called a Möbius transformation, which has the nice property of mapping circles to circles in $\Bbb C\cup\{\infty\}$. In particular, then, each $M(z)$ should be a circle or a line in $\Bbb C$.

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