Find the modulus of $|z-5|/|1-3z|$ when z is given

Question

If $z = 3-2i$ then find $$\frac { \left| z-5 \right| }{ \left| 1-3z \right| } $$ I've substituted z by $|z|^2/z$ conjugate but still cant figure out what to do,

Thanks in advance

Answer 1

$$\frac { \left| z-5 \right| }{ \left| 1-3z \right| } =\frac { \left| 3-2i-5 \right| }{ \left| 1-9+6i \right| } =\frac { \left| -2-2i \right| }{ \left| -8+6i \right| } =\frac { \sqrt { { 2 }^{ 2 }+{ 2 }^{ 2 } } }{ \sqrt { { \left( -8 \right) }^{ 2 }+{ 6 }^{ 2 } } } =\frac { 2\sqrt { 2 } }{ 10 } =\frac { \sqrt { 2 } }{ 5 } $$

Answer 2

Notice, when $z_1\space\wedge\space z_2\in\mathbb{C}$:

$$\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}=\frac{\left|\Re[z_1]+\Im[z_1]i\right|}{\left|\Re[z_2]+\Im[z_2]i\right|}=\frac{\sqrt{\Re[z_1]+\Im[z_1]}}{\sqrt{\Re^2[z_2]+\Im^2[z_2]}}$$

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Another way:

$$\frac{(3-2i)-5}{1-3(3-2i)}=\frac{3-2i-5}{1-9+6i}=\frac{-2-2i}{-8+6i}=$$ $$\frac{(-2-2i)(-8-6i)}{(-8+6i)(-8-6i)}=\frac{16+12i+16i+12ii}{8^2+6^2}=$$ $$\frac{16-12+28i}{100}=\frac{4+28i}{100}\to\left|\frac{4+28i}{100}\right|=\frac{\sqrt{4^2+28^2}}{100}=\frac{\sqrt{2}}{5}$$