# Image of a unit disk under a Möbius transformation

I have to find the image of the disk $$|z| < 1$$ under the Möbius Transformation $$w(z) = \frac{iz-i}{z+1}$$. This is how I approached it:

Since every Möbius function is determined uniquely by its action on three points. I took three points on the boundary of the disk (i.e. unit circle) $$i, 1$$ and $$-i$$ whose images were $$-i, 0$$ and $$1$$ respectively.

Since Möbius transformations are conformal and the unit disk was on the right hand side of the three points, the image must be on the right hand side of the three images. This implies that the image must be the 4th quadrant, excluding the positive real axis and negative imaginary axis.

Is this a valid solution? I was trying to attempt a lot of exercises like this during self-learning complex functions. Is my approach rigorous and valid enough?

• Mobius transformations map circles to lines or circles, so the image of the unit disk can be either a disk or a half-plane, but never a quadrant. – dxiv Jun 6 '17 at 3:45
• Your mistake seems to be your calculation of $w(i)$. – Lubin Jun 6 '17 at 3:59

You want to find the image of $$\vert z\vert<1$$ under $$T(z)=\frac{iz-i}{z+1}=\frac{i(z-1)}{z+1}$$.

Take 3 boundary points: $$1,i,-1$$ and

note that

• $$T(1)=0$$

• $$T(i)=-1$$

• $$T(-1)=\infty$$

Also observe the direction of the points, first is $$1$$ then $$i$$ and then $$-1;$$ and the interior is in the LHS, thus following the same direction of the points $$0,-1,\infty$$, the interior should be also in the LHS.

Alternatively you could take an arbitrary point of the interior of $$|z|<1,$$ say $$0$$, and $$T(0)=-i$$ tells you where the image is.

Thus the image of $$|z|<1$$ is the lower half-plane.