Mapping of Complex function Let $w = f(w)=(3+4i)z- 2 +i$  Find the images of the disk $|z-1|<1$ and the half-plane $Im(z)>1$
For the first mapping of the disk I attempted the following :
$$z = \frac{w+2-i}{3+4i}$$ which then gives us
$$|\frac{w+2-i}{3+4i} - 1| < 1$$ 
$$|w-(1+5i)| < |3+4i|$$
$$|w-(1+5i)| < 5$$ And this is just a circle with origin (1,5) with a radius of 5
For the second image I did the following :
Let $z=x+iy$ thus $$w =(3+4i)(x+iy)-2+i=(3x-4y-2)+i(3y+4x+1)$$
Then $y=1$ gives $u=3x-2$ and $v=4+4x$ which leads to the line $u=\frac{3}{4}v-5$
Are these mappings correct ?
 A: Since these are both examples of Mobius Transforms, the images of generalized circles (aka circles and lines) are generalized circles, which confirms already your answers are not unexpected.
Stronger, your function is of the form $az+b$ for complex $a$ and $b.$ This is merely a superposition of two actions: 


*

*$az$ which scales and rotates the value $z$ by a fixed amount

*$z+b,$ which shifts the parameter $z$ in the plane.


Geometrically, we cannot convert circles to lines or lines to circles via such actions, so circles must map to circles and lines must map to lines.

For the first case, we have a circle: to find the circle which is its image, we first find the center of the image:
$$f(1) = 1+5i$$ 
and then figure out where a point on the edge is mapped to:
 $$f(0) = -2+i$$
Since we have $$\sqrt{(-2 - 1)^2 + (1 - 5)^2} = 5$$ we conclude we get a circle centered at $1 + 5i$ of radius $5$, which we might write as
$$\bbox[6px,border:2px solid]{|w-(1+5i)| < 5}$$

For the second case, note the geometry forces the image of a half plane to remain a half plane, with the boundary of our image being the image of our initial boundary. With this in mind, let us consider the images under $f$ of two points on our initial boundary, the line $\operatorname{Im}(z) = 1:$
$$f(i) = -6 + 4i$$ $$f(i+1) = -3 + 8i$$ As mentioned prior, the image must be the line through these two points, which in the coordinates on $\mathbb{R^2}$ is simply $$y = \tfrac{4}{3}x + 12$$ We must now determine which side of this line our half plane maps onto. To this end, note
$$f(2i) = -10 + 7i$$
which lies above our line. Thus our image is the set
$$\bbox[5px,border:2px solid]{\operatorname{Im}(z) > \tfrac{4}{3}\operatorname{Re}(z) + 12}$$
