<s>[INST] Find the range of change number $c=\frac{\text{Im } z}{\text{Re } z}$ if $|z-3-4i|\le 1$ I am working on the following problem:
Find the range of change of the numbers $c=\frac{\text{Im }z}{\text{Re }z}$ if $|z-3-4i|\le 1$
This is what I did:
Letting $z=x+iy$, I can write from $|z-3-4i|\le 1$ $$(x-3)^2+(y-4)^2\le 1$$ From that it can be written $$2\le x\le 4$$ and $$3\le y\le 5$$ Therefore $$\frac{3}{4}\le c\le\frac{5}{2}$$
Is there anything wrong with my work? Any help would be greatly appreciated.
 [/INST] We have a circle centered at $(3,4)$ of radius $1.$
construct the line between the origin and the center of the circle.
Let $\theta$ be the angle that this line forms with the x axis.
$\tan \theta = \frac 43$
Construct the line between the origin and the tangents.
Let $\phi$ be the angles between each tangent and the line to the center of the circle.
$\sin \phi = \frac 15\\
\cos \phi = \sqrt {1-\sin^2\phi} = \frac {\sqrt {24}}{5}\\  
\tan \phi = \frac {\sqrt 6}{12}$
$c = [\tan (\theta - \phi), \tan (\theta + \phi)]$
$\tan (\theta + \phi) = \frac {tan \theta + \tan \phi}{1-\tan\theta \tan\phi} = \frac {48+3\sqrt{6}}{36 - 4\sqrt {6}} = \frac {6+\sqrt 6}{4}\\
\tan (\theta - \phi) = \frac {tan \theta - \tan \phi}{1+\tan\theta \tan\phi} = \frac {48-3\sqrt 6}{36+4\sqrt 6}= \frac {6-\sqrt 6}{4}$
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