Show Möbius transformation which maps $\Im(z)>0$ to itself iff

$$ f(z)= \frac{az+b}{cz+d}\,,\,\,ad-bc>0$$ and $a,b,c,d$ are real.

  • 1
    $\begingroup$ This question does not show much effort on the part of the asker. $\endgroup$ – Ben Millwood Mar 26 '13 at 2:29
  • $\begingroup$ What do you call $J(z)>0$ $\endgroup$ – superAnnoyingUser Mar 27 '13 at 20:38
  • $\begingroup$ You should word the question in a way that makes sense? $\endgroup$ – superAnnoyingUser Mar 27 '13 at 20:42


$$ Im((az + b)(c\bar{z} + d)) = (ad - bc)Im(z) $$

Edit for more detail as requested:

Multiplying $f(z)$ by $\dfrac{c\bar{z} + d}{c\bar{z} + d}$, we are left with:

$$\dfrac{(az + b)(c \bar{z} + d)}{(cz + d)(c \bar{z} + d)} = \dfrac{(ac(z\bar{z}) + bd) + (bc \bar{z} + adz)}{(cz + d)\overline{(cz + d)}}$$

Now both $z \bar{z}$ and $z + \bar{z}$ are real, so the imaginary part of this expression is:

$$ Q = \dfrac{ad - bc}{(cz + d)\overline{(cz + d)}} Im(z) $$

Because $w \bar{w} \geq 0$ for all $w$, the sign of $\frac{Q}{Im(z)}$ must be the same as the sign of $ad - bc$. Therefore they have the same sign if $ad - bc > 0$

Second part of the question is that this map is onto, ie) every $z$ value has a preimage $w$, ie) $f(w) = z$

To see this is the case, imagine a vector in $\mathbb{C}^2$ as representing a ratio of two complex numbers. Then $\left( \begin{array}{c} z \\ 1 \end{array} \right)$ represents the ratio $\frac{z}{1}$. Then $f(z)$ is represented by the matrix equation:

$$ f \left( \begin{array}{c} z \\ 1 \end{array} \right) = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \left( \begin{array}{c} z \\ 1 \end{array} \right)$$


$$ f = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) $$

Because the determinant $ad - bc > 0$ is non-zero, this map is invertible. This means that for any $ Z = \left( \begin{array}{c} z_1 \\ z_2 \end{array} \right)$, there is a $W = \left( \begin{array}{c} w_1 \\ w_2 \end{array} \right)$ such that $f(W) = Z$.

Dividing by $w_2$, we obtain the same ratio as before but with $W$ in the form that we know. If $w_2$ is $0$, then $z_2$ is $0$ so this will never be a problem.

  • $\begingroup$ That \Im is awful, how do you make a regular Im without \text ? $\endgroup$ – muzzlator Mar 19 '13 at 17:43
  • $\begingroup$ Both Im, Re are really ugly and pretty non-standard. I don't know why this is so... $\endgroup$ – DonAntonio Mar 19 '13 at 17:57
  • $\begingroup$ @muzzlator we can use \operatorname{Im} $\operatorname{Im}(z)$ or \mathrm{Im} $\mathrm{Im}(z)$. $\endgroup$ – Américo Tavares Mar 19 '13 at 19:06
  • $\begingroup$ Thanks muzzlator. Could you explain your hint? Appreciate it. $\endgroup$ – sarah Mar 19 '13 at 19:18
  • $\begingroup$ @maria Multiply $f(z)$ by $\dfrac{c \bar{z} + d}{c \bar{z} + d}$ and use the fact $z \bar{z}$ is real. $\endgroup$ – muzzlator Mar 20 '13 at 6:23

Möbius transformations map lines and circles to lines and circles. They are also continuous functions of the Riemann sphere. In some sense, lines are circles in the geometry of the Riemann sphere.

This information alone is enough to give a strong qualitative description of those Möbius transformations that map the upper half-plane to itself: they must map the real line (plus $\infty$) to itself, because that is the boundary of the upper half-plane.

Furthermore, it is orientation preserving; as you progress along the real line in the positive direction, the upper half-plane is on your left. This relationship is preserved by any Möbius transformation. So we require that the real line is not only mapped to itself, but it is mapped to itself in the same direction.

From the first observation, it should be straightforward to determine the coefficients are real. The second observation gives us the sign condition. (although depending on your knowledge, it might be easier to just look at the sign on the imaginary part of $f(i)$)

  • $\begingroup$ Why is it just enough to just look at f(i)? how does Im(f(i))>0 imply that the upper plane gets mapped to the upper plane? thanks $\endgroup$ – sarah Mar 22 '13 at 3:37
  • $\begingroup$ Without crossing the real line (plus $\infty$), you can get from any point in the upper half-plane to any other, but not to any point in the lower half-plane. Each entire half plane has to lie on one "side" of the real line (plus $\infty$) or the other. So picking one point is enough to find out which side a half-plane goes to. $\endgroup$ – Hurkyl Mar 22 '13 at 4:09
  • $\begingroup$ Why should the real line go to itself? Could you please explain? Thank you. $\endgroup$ – WhySee Mar 19 '18 at 11:04

Let $H_{+}=\{z\in \mathbb C: \Im (z)>0\}$ and let $H_{-}=\{z\in \mathbb C: \Im (z)<0\}$. Since $H_{+}$ is connected, therefore, $f(H_{+})$ is connected. Thus $f(H_{+})$ can not intersect both $H_{+}$ and $H_{-}$. Therefore, $f(H_{+})\subset H_{+}$ or $f(H_{+})\subset H_{-}$.

Since $f(i)=\dfrac{ai+b}{ci+d}=\dfrac{ac+bd+i(ad-bc)}{c^2+d^2} \in H_{+}$. Hence $f$ maps $H_{+}$ onto $H_{+}$


Denote $z:=u+iv$. Then (a little bit tedious) calculation gives $$ \operatorname{Im} f(u+iv)=\frac{v(ad-bc)}{(cu+d)^2+c^2v^2}. $$ If $c=0$ then $$ \operatorname{Im} f(u+iv)=\frac{a}{d}v. $$ If you want that it to be positive for every $v>0$ then the necessary and sufficient condition is $ad>0$, that is, $ad=ad-bc>0$.

If $c\neq0$ then $(cu+d)^2+c^2v^2>0$ and the necessary and sufficient condition is $ad-bc>0$.


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