Show that by means of the transformation $w=\frac 1z$ the circle $C$ given by $|z-3|=5$ is mapped into the circle $|w+\frac{3}{16}|=\frac{5}{16}$

Question

Show that by means of the transformation $w=\frac 1z$ the circle C given by $|z-3|=5$ is mapped into the circle $\left|w+\frac{3}{16}\right|=\frac{5}{16}$

My try: $$\begin{align}\\ &w=\frac 1z\\ &\implies 3w=\frac 3z\\ &\implies1-3w=1-\frac 3z=\frac{z-3}{z}\\ &\implies|1-3w|=\left|\frac{z-3}{z}\right|\\ &\implies|1-3w|=|z-3|\left|\frac 1z\right|\\ &\implies|1-3w|=5|w|\\ \end{align}\\$$ But, I can't understand what to do next and how can I proof that. Any suggestions? or, you can add an answer.

Answer 1

$$|z-3|=5.$$ Let $\displaystyle w= \frac{1}{z}$ and let $w^\star$ be the conjugate of $w$.

$$\begin{aligned} \left(\frac{1}{w} -3 \right) \left(\frac{1}{w^\star} -3 \right) &= 25 \\ (1 -3 w^\star) \left( 1 - 3w \right) &= 25 w w^\star\\ 1-3(w+w^\star) + 9 w w^\star &= 25 w w^\star \\ 16 w w^\star + 3(w + w^\star) -1 &= 0\\ w w^\star + \frac{3}{16} (w + w^\star) &= \frac{1}{16}\\ \left( w + \frac{3}{16}\right) \left( w^\star + \frac{3}{16} \right) &= \frac{9}{16^2} + \frac{1}{16} = \left( \frac{5}{16}\right)^2\\ \left| w + \frac{3}{16} \right| &= \frac{5}{16}. \end{aligned}$$

Answer 2

We need to prove that if $|z-3|=5$ so $$\left|\frac{1}{z}+\frac{3}{16}\right|=\frac{5}{16}.$$ Let $z=x+yi,$ where $x$ and $y$ are reals.

Thus, $$(x-3)^2+y^2=25$$ and we need to prove that: $$|3z+16|=5|z|$$ or $$(3x+16)^2+9y^2=25(x^2+y^2)$$ or $$16(x^2+y^2)=96x+256$$ or $$x^2-6x+y^2=16$$ or $$(x-3)^2+y^2=25$$ and we are done!