How do you find the min/max values of $Arg(z)$ for {$z:|z-(4+3i)|=2$}?

Question

How do you find the min/max values of $Arg(z)$ for {$z:|z-(4+3i)|=2$}?

I have drawn a diagram with a circle centre $(4+3i)$ and radius = $2$. I think to find the min/max values of $Arg(z)$ I need to find the points of tangency? I don't understand how to do it only with the diagram and any hints would be much appreciated!

The min possible value of $|z|$ is $3$, and max value of $|z|$ is $7$.

Also, I am on a chapter has more simple questions like this, so I'm looking for a not too complicated way of solving this :)

Answer 1

In the following picture, the points you require are $C$ and $D$, since any point above the line $AC$ (and hence, with slope more than that of $AC$) and below the line $AD$ (and hence, with slope less than that of $AD$) will not lie on the circle. Since, the whole circle is in the first quadrant, the slope of the line joining a point $D$ on the circle and the origin is same as $\tan\theta$ where $\theta$ is the argument of the complex number $D(x,y):=x+\iota y$.

According to the problem, the center of the circle is $B(4,3)$ and radius is $BD=BC=2$

Now, what you need is $\angle DAE$ and $\angle CAE$ ($BE \perp AE$) $$\begin{aligned} \angle DAE &= \angle BAE - \angle BAD \\ & = \sin^{-1}\dfrac{BE}{BA} - \sin^{-1} \dfrac{DB}{BA}\\ & = \sin^{-1} \dfrac35-\sin^{-1} \dfrac25 \ (\because BA = \sqrt{4^2+3^2}=5) \\ &=(0.6435 - 0.4115)\text{rad}=0.232\text{ rad}\\ \text{and } \ \angle CAE &=\angle BAE+\angle CAB \\&= \sin^{-1}\dfrac{BE}{BA} +\sin^{-1} \dfrac{CB}{BA}\\&=\sin^{-1} \dfrac35+\sin^{-1} \dfrac25\\ &=(0.6435 +0.4115)\text{rad}=1.055\text{ rad}\end{aligned}$$

Answer 2

Let $y=mx$ be the equation of a line tangent to the circle $(x-4)^2 + (y-3)^2 =4$. Then substituting the first into the second, we get

$$ \begin{align} (x-4)^2 + (mx-3)^2 &= 4 \\ (m^2 + 1) \, x^2 - (6m+8) \, x +21 &= 0 \end{align} $$

Since the line is tangent to the circle, the discriminant of the above must be zero.

$$ \begin{align} (6m+8)^2 - 4 (m^2 + 1)(21) &= 0 \\ 12m^2 - 24m + 5 &= 0 \\ m &= \frac{6 \pm \sqrt{21}}{6} \end{align} $$

Finally, we have $\tan \theta = m$, so

$$\theta_1 = \tan^{-1} \left( \frac{6 - \sqrt{21}}{6} \right) = 0.232$$

and

$$\theta_2 = \tan^{-1} \left( \frac{6 + \sqrt{21}}{6} \right) = 1.055$$